<div id="pf1" class="pf w0 h0" data-page-no="1"><div class="pc pc1 w0 h0"><img class="bi x0 y0 w1 h1" alt="" src="https://files.passeidireto.com/ae95c9c7-2446-4b64-84fc-96c538ce870f/bg1.png"><div class="c x0 y1 w2 h2"><div class="t m0 x1 h3 y2 ff1 fs0 fc0 sc0 ls24 ws0">EA<span class="_0 blank"></span>E0<span class="_0 blank"></span>207<span class="_0 blank"></span>:<span class="_1 blank"> </span>M<span class="_0 blank"></span>ate<span class="_0 blank"></span>mát<span class="_0 blank"></span>ica A<span class="_0 blank"></span>pli<span class="_0 blank"></span>cad<span class="_0 blank"></span>a à Eco<span class="_0 blank"></span>no<span class="_0 blank"></span>mia</div><div class="t m0 x2 h4 y3 ff1 fs1 fc0 sc0 ls24 ws1">Aula 18:<span class="_2 blank"> </span>Equações Diferenciais de Primeira Ordem (Cont.)</div><div class="t m0 x3 h4 y4 ff1 fs1 fc1 sc0 ls24 ws2">Ma<span class="_0 blank"></span>rcos Y. Nakaguma</div><div class="t m0 x4 h4 y5 ff1 fs1 fc1 sc0 ls24 ws3">06/10/2017</div><div class="t m0 x5 h5 y6 ff1 fs2 fc1 sc0 ls24">1</div></div><div class="c x0 y7 w2 h2"><div class="t m0 x6 h3 y8 ff1 fs0 fc0 sc0 ls24 ws4">Eq<span class="_0 blank"></span>ua<span class="_0 blank"></span>ções D<span class="_0 blank"></span>ife<span class="_0 blank"></span>ren<span class="_0 blank"></span>ciai<span class="_0 blank"></span>s Lin<span class="_0 blank"></span>ea<span class="_0 blank"></span>re<span class="_0 blank"></span>s de P<span class="_0 blank"></span>rim<span class="_0 blank"></span>eir<span class="_0 blank"></span>a-Or<span class="_0 blank"></span>dem</div><div class="t m0 x7 h4 y9 ff1 fs1 fc1 sc0 ls24 ws1">As equações diferenciais linea<span class="_0 blank"></span>res de primeira-o<span class="_0 blank"></span>rdem com <span class="fc2 ws5">co e\u2026cientes</span></div><div class="t m0 x7 h4 ya ff1 fs1 fc2 sc0 ls24 ws6">constantes <span class="fc1 ws7 v0">p<span class="_3 blank"> </span>ossuem a seguinte fo<span class="_0 blank"></span>rma:</span></div><div class="t m0 x8 h6 yb ff2 fs1 fc1 sc0 ls24">dy</div><div class="t m0 x8 h7 yc ff2 fs1 fc1 sc0 ls24 ws8">dt <span class="ff3 fs3 ls0 v1">=</span><span class="fc2 ls1 v1">a</span><span class="ls2 v1">y<span class="ff3 fs3 ls3">+</span><span class="ls4">b<span class="ff3 fs3 ls5 v0">(</span><span class="ls6">t</span></span></span><span class="ff3 fs3 v1">)</span></div><div class="t m0 x7 h4 yd ff1 fs1 fc1 sc0 ls24 ws1">Quanto à terminologia, dizemos que:</div><div class="t m0 x9 h8 ye ff2 fs4 fc0 sc0 ls7">i<span class="ff4 ls8">.<span class="ff1 fc1 ls24 ws9">A equação é <span class="fc2 wsa">autônoma </span><span class="wsb">se <span class="ff2 ls9">b<span class="ff3 fs5 lsa v0">(</span><span class="lsb">t<span class="ff3 fs5 lsc v0">)<span class="lsd v0">=</span></span><span class="lse">b</span></span></span><span class="wsc">, i.e.:</span></span></span></span></div><div class="t m0 xa h9 yf ff2 fs4 fc1 sc0 ls24 wsd">dy</div><div class="t m0 xa ha y10 ff2 fs4 fc1 sc0 ls24 wse">dt <span class="ff3 fs5 lsf v2">=</span><span class="fc2 ls10 v2">a</span><span class="ls11 v2">y<span class="ff3 fs5 ls12">+</span></span><span class="fc2 v2">b</span></div><div class="t m0 xb h8 y11 ff2 fs4 fc0 sc0 ls24 wsf">ii <span class="ff4 ls13">.</span><span class="ff1 fc1 ws9">A equação é <span class="fc2 ws10">homogênea </span><span class="wsb">se <span class="ff2 ls14">b<span class="ff3 fs5 ls15 v0">(</span><span class="lsb">t<span class="ff3 fs5 lsc v0">)<span class="lsf v0">=</span></span></span></span><span class="wsc">0, i.e.:</span></span></span></div><div class="t m0 xc h9 y12 ff2 fs4 fc1 sc0 ls24">dy</div><div class="t m0 xc hb y13 ff2 fs4 fc1 sc0 ls24 wse">dt <span class="ff3 fs5 ls16 v2">=</span><span class="fc2 ls10 v2">a</span><span class="v2">y</span></div><div class="t m0 x5 h5 y14 ff1 fs2 fc1 sc0 ls24">2</div></div><div class="c x0 y15 w2 h2"><div class="t m0 x6 h3 y16 ff1 fs0 fc0 sc0 ls24 ws4">Eq<span class="_0 blank"></span>ua<span class="_0 blank"></span>ções D<span class="_0 blank"></span>ife<span class="_0 blank"></span>ren<span class="_0 blank"></span>ciai<span class="_0 blank"></span>s Lin<span class="_0 blank"></span>ea<span class="_0 blank"></span>re<span class="_0 blank"></span>s de P<span class="_0 blank"></span>rim<span class="_0 blank"></span>eir<span class="_0 blank"></span>a-Or<span class="_0 blank"></span>dem</div><div class="t m0 x7 h4 y17 ff1 fs1 fc1 sc0 ls24 ws1">Na aula passada, estudamos as equações diferenciais lineares de</div><div class="t m0 x7 h4 y18 ff1 fs1 fc1 sc0 ls24 ws11">p<span class="_0 blank"></span>rimeira-ordem com <span class="fc2 ws5 v0">coe\u2026cientes<span class="_4 blank"> </span>constantes<span class="_4 blank"> </span><span class="fc1 ws12">do<span class="_5 blank"> </span>seguinte<span class="_5 blank"> </span>tip o:</span></span></div><div class="t m0 x8 h6 y19 ff2 fs1 fc1 sc0 ls24">dy</div><div class="t m0 x8 hc y1a ff2 fs1 fc1 sc0 ls24 ws8">dt <span class="ff3 fs3 ls0 v1">=</span><span class="fc2 ls1 v1">a</span><span class="ls2 v1">y<span class="ff3 fs3 ls3">+</span><span class="ls4">b<span class="ff3 fs3 ls5 v0">(</span><span class="ls6">t</span></span></span><span class="ff3 fs3 v1">)</span></div><div class="t m0 x7 h4 y1b ff1 fs1 fc1 sc0 ls24 ws13">Neste caso, vimos que a <span class="fc2 ws14">solução geral </span><span class="ws7">é dada p<span class="_3 blank"> </span>o<span class="_0 blank"></span>r:</span></div><div class="t m0 xd hd y1c ff2 fs1 fc1 sc0 ls17">y<span class="ff3 fs3 ls18 v0">(</span><span class="ls6">t<span class="ff3 fs3 ls19 v0">)<span class="ls1a v0">=</span></span><span class="ls24 ws15">k<span class="_0 blank"></span>e <span class="fs2 ws16 v3">a t<span class="_5 blank"> </span></span><span class="ff3 fs3 ls3">+</span><span class="ls1b">e</span><span class="fs2 ws16 v3">a t<span class="_5 blank"> </span></span><span class="ff5 fs3 ls1c v4">R</span><span class="ls1d">e<span class="ff6 fs6 ls1e v3">\ue000</span></span><span class="fs2 ws16 v3">at<span class="_6 blank"> </span></span><span class="ls1f">b<span class="ff3 fs3 ls5 v0">(</span><span class="ls6">t<span class="ff3 fs3 ls20 v0">)</span></span></span><span class="ws17">dt <span class="ff4">,</span></span></span></span></div><div class="t m0 x7 he y1d ff1 fs1 fc1 sc0 ls24 ws18">com <span class="ff2 ls21">k<span class="ff6 fs3 ls22">2</span><span class="ff7 ls23">R</span></span><span class="ff4">.</span></div><div class="t m0 x5 h5 y1e ff1 fs2 fc1 sc0 ls24">3</div></div></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div> <div id="pf2" class="pf w0 h0" data-page-no="2"><div class="pc pc2 w0 h0"><img class="bi x0 y0 w1 h1" alt="" src="https://files.passeidireto.com/ae95c9c7-2446-4b64-84fc-96c538ce870f/bg2.png"><div class="c x0 y1 w2 h2"><div class="t m0 x6 h3 y1f ff1 fs0 fc0 sc0 ls24 ws1a">Eq<span class="_0 blank"></span>.<span class="_1 blank"> </span>D<span class="_0 blank"></span>ifer<span class="_0 blank"></span>enc<span class="_0 blank"></span>iais L<span class="_0 blank"></span>in<span class="_0 blank"></span>ea<span class="_0 blank"></span>res A<span class="_0 blank"></span>ut<span class="_0 blank"></span>ôno<span class="_0 blank"></span>ma<span class="_0 blank"></span>s de P<span class="_0 blank"></span>rim<span class="_0 blank"></span>eir<span class="_0 blank"></span>a-Or<span class="_0 blank"></span>dem</div><div class="t m0 x7 h4 y20 ff1 fs1 fc1 sc0 ls24 ws13">Observe que no caso de uma equação diferencial <span class="fc2 ws1b">autônoma </span><span class="ws12">do<span class="_5 blank"> </span>tip o:</span></div><div class="t m0 xe h6 y21 ff2 fs1 fc1 sc0 ls24">dy</div><div class="t m0 xe hf y22 ff2 fs1 fc1 sc0 ls24 ws8">dt <span class="ff3 fs3 ls25 v1">=</span><span class="fc2 ls26 v1">a</span><span class="ls2 v1">y<span class="ff3 fs3 ls27">+</span><span class="fc2 ls28">b<span class="ff4 fc1 ls29">,<span class="ff1 ls24 ws18">com <span class="ff2 ls2a">a<span class="ff6 fs3 ls2b">6<span class="ff3 ls2c">=</span></span></span>0</span></span></span></span></div><div class="t m0 x7 h4 y23 ff1 fs1 fc1 sc0 ls24 ws1c">a solução seria dada p<span class="_3 blank"> </span>o<span class="_0 blank"></span>r:</div><div class="t m0 xf h10 y24 ff2 fs1 fc1 sc0 ls2d">y<span class="ff3 fs3 ls5 v0">(</span><span class="ls6 v0">t<span class="ff3 fs3 ls19 v0">)<span class="ls1a v0">=</span></span><span class="ls24 ws1d">k<span class="_0 blank"></span>e <span class="fs2 ws1e v5">a t<span class="_5 blank"> </span></span><span class="ff3 fs3 ls3">+</span><span class="ls1b">e</span><span class="fs2 ws1e v5">a t<span class="_5 blank"> </span></span><span class="ff5 fs7 ls2e v6">Z</span><span class="ls2f v0">e<span class="ff6 fs6 ls30 v5">\ue000</span></span><span class="fs2 ws16 v5">at<span class="_6 blank"> </span></span><span class="ws1f v0">b dt</span></span></span></div><div class="t m0 x7 h4 y25 ff1 fs1 fc1 sc0 ls24 ws20">Assim, temos que:</div><div class="t m0 x10 h11 y26 ff2 fs1 fc1 sc0 ls31">y<span class="ff3 fs3 ls32 v0">(</span><span class="ls33 v0">t<span class="ff3 fs3 ls34 v0">)<span class="ls35 v0">=</span></span><span class="ls24 ws3">k<span class="ff1 fs2 ls36 v7">1</span><span class="ls37">e</span><span class="fs2 ws16 v5">at<span class="_4 blank"> </span></span><span class="ff3 fs3 ls3">+</span><span class="ws12">be<span class="_7 blank"> </span><span class="fs2 ws16 v5">at<span class="_5 blank"> </span></span><span class="ff5 ls38 v8">\ue012</span><span class="ff6 fs3 ls39">\ue000</span><span class="ff1 v1">1</span></span></span></span></div><div class="t m0 x11 h12 y27 ff2 fs1 fc1 sc0 ls3a">a<span class="ls2f v1">e</span><span class="ff6 fs6 ls30 v9">\ue000</span><span class="fs2 ls24 ws16 v9">at<span class="_6 blank"> </span></span><span class="ff5 ls24 va">\ue013</span></div><div class="t m0 x7 h4 y28 ff1 fs1 fc1 sc0 ls24 ws21">o que implica que:</div><div class="t m0 x4 h13 y29 ff2 fs1 fc1 sc0 ls3b">y<span class="ff3 fs3 ls32 v0">(</span><span class="ls33 v0">t<span class="ff3 fs3 ls3c v0">)<span class="ls3d v0">=</span></span><span class="ls24 ws22">k<span class="_0 blank"></span>e <span class="fs2 ws16 v5">a t<span class="_5 blank"> </span></span><span class="ff6 fs3 ls3e">\ue000</span><span class="v1">b</span></span></span></div><div class="t m0 x12 h6 y2a ff2 fs1 fc1 sc0 ls24">a</div><div class="t m0 x5 h5 y2b ff1 fs2 fc1 sc0 ls24">4</div></div><div class="c x0 y7 w2 h2"><div class="t m0 x6 h3 y8 ff1 fs0 fc0 sc0 ls24 ws23">Eq<span class="_0 blank"></span>.<span class="_1 blank"> </span>D<span class="_0 blank"></span>ifer<span class="_0 blank"></span>enc<span class="_0 blank"></span>iais L<span class="_0 blank"></span>in<span class="_0 blank"></span>ea<span class="_0 blank"></span>re<span class="_0 blank"></span>s Au<span class="_0 blank"></span>tôn<span class="_0 blank"></span>oma<span class="_0 blank"></span>s de P<span class="_0 blank"></span>rim<span class="_0 blank"></span>eir<span class="_0 blank"></span>a-O<span class="_0 blank"></span>rde<span class="_0 blank"></span>m</div><div class="t m0 x7 h4 y2c ff1 fs1 fc1 sc0 ls24 ws13">P<span class="_0 blank"></span>ortanto, a <span class="fc2 ws24">solução geral </span><span class="ws12">da<span class="_5 blank"> </span>equação<span class="_5 blank"> </span>p o<span class="_8 blank"> </span>de<span class="_5 blank"> </span>ser<span class="_4 blank"> </span>exp<span class="_0 blank"></span>ressa<span class="_4 blank"> </span>como:</span></div><div class="t m0 x4 h14 y2d ff2 fs1 fc1 sc0 ls3f">y<span class="ff3 fs3 ls5 v0">(</span><span class="ls6">t<span class="ff3 fs3 ls34 v0">)<span class="ls40 v0">=</span></span><span class="ls24 ws1d">k<span class="_0 blank"></span>e <span class="fs2 ws16 v3">a t<span class="_5 blank"> </span></span><span class="ff6 fs3 ls3e">\ue000</span><span class="fs2 v5">b</span></span></span></div><div class="t m0 x12 h15 y2e ff2 fs2 fc1 sc0 ls24">a</div><div class="t m0 x7 h16 y2f ff1 fs1 fc1 sc0 ls24 ws1">Dada uma condição inicial <span class="ff2 ls17">y<span class="ff3 fs3 ls5 v0">(</span></span><span class="ls41">0<span class="ff3 fs3 ls19 v0">)<span class="ls1a v0">=</span></span><span class="ff2 ls42">y</span><span class="fs2 ls43 v7">0</span></span>, devemos ter que:</div><div class="t m0 xe h17 y30 ff2 fs1 fc1 sc0 ls24 ws3">y<span class="ff1 fs2 ls44 v7">0</span><span class="ff3 fs3 ls25 v0">=</span><span class="ls45 v0">k<span class="ff6 fs3 ls46">\ue000</span></span><span class="v1">b</span></div><div class="t m0 x13 h18 y31 ff2 fs1 fc1 sc0 ls47">a<span class="ff6 fs3 ls48 v1">)</span><span class="ls49 v1">k<span class="ff3 fs3 ls25">=</span><span class="ls42">y<span class="ff1 fs2 ls4a v7">0</span><span class="ff3 fs3 ls46">+</span><span class="ls24 v1">b</span></span></span></div><div class="t m0 x14 h6 y31 ff2 fs1 fc1 sc0 ls24">a</div><div class="t m0 x7 h4 y32 ff1 fs1 fc1 sc0 ls24 ws13">P<span class="_0 blank"></span>ortanto, a <span class="fc2 v0">solução pa<span class="_0 blank"></span>rticular <span class="fc1 ws7">é dada po<span class="_0 blank"></span>r:</span></span></div><div class="t m0 x15 h19 y33 ff2 fs1 fc1 sc0 ls4b">y<span class="ff3 fs3 ls32 v0">(</span><span class="ls6">t<span class="ff3 fs3 ls19 v0">)<span class="ls4c v0">=</span></span><span class="ff5 ls4d v4">\ue000</span><span class="ls24 ws3 v0">y<span class="ff1 fs2 ls4e v7">0</span></span><span class="ff3 fs3 ls3e">+</span><span class="fs2 ls24 v5">b</span></span></div><div class="t m0 x16 h1a y34 ff2 fs2 fc1 sc0 ls4f">a<span class="ff5 fs1 ls50 vb">\ue001</span><span class="fs1 ls37 v3">e</span><span class="ls24 ws16 v1">at<span class="_4 blank"> </span></span><span class="ff6 fs3 ls46 v3">\ue000</span><span class="ls24 vc">b</span></div><div class="t m0 x17 h15 y34 ff2 fs2 fc1 sc0 ls24">a</div><div class="t m0 x7 h14 y35 ff1 fs1 fc1 sc0 ls24 ws25">Note que se <span class="ff2 ws3">y</span><span class="fs2 ls51 v7">0</span><span class="ff3 fs3 ls52">=<span class="ff6 ls53">\ue000</span></span><span class="ff2 fs2 v5">b</span></div><div class="t m0 xf h1b y36 ff2 fs2 fc1 sc0 ls54">a<span class="ff4 fs1 ls55 v3">,<span class="ff1 ls24 ws26">então <span class="ff2 ls56">y<span class="ff3 fs3 ls18 v0">(</span><span class="ls33 v0">t<span class="ff3 fs3 ls3c v0">)<span class="ls57 v0">=<span class="ff6 ls58">\ue000</span></span></span></span></span></span></span><span class="ls24 vc">b</span></div><div class="t m0 x18 h1c y36 ff2 fs2 fc1 sc0 ls59">a<span class="ff1 fs1 ls24 ws5 v3">pa<span class="_0 blank"></span>ra<span class="_5 blank"> </span>to do<span class="_5 blank"> </span><span class="ff2 ls5a">t</span><span class="ws27">.<span class="_2 blank"> </span>Assim, dizemos</span></span></div><div class="t m0 x7 h1d y37 ff1 fs1 fc1 sc0 ls24 ws28">que <span class="ff2 ls5b v0">y<span class="ff3 fs3 ls57">=<span class="ff6 ls39">\ue000</span></span><span class="fs2 ls24 v5">b</span></span></div><div class="t m0 x19 h1c y38 ff2 fs2 fc1 sc0 ls5c">a<span class="ff1 fs1 ls24 ws29 v3">é um <span class="fc2 ws2a v0">p onto<span class="_5 blank"> </span>de<span class="_4 blank"> </span>equilíb<span class="_0 blank"></span>rio<span class="_4 blank"> </span><span class="fc1 ws21">ou uma <span class="fc2">solução estacionária</span>.</span></span></span></div><div class="t m0 x5 h5 y39 ff1 fs2 fc1 sc0 ls24">5</div></div><div class="c x0 y15 w2 h2"><div class="t m0 x6 h3 y16 ff1 fs0 fc0 sc0 ls24 ws23">Eq<span class="_0 blank"></span>.<span class="_1 blank"> </span>D<span class="_0 blank"></span>ifer<span class="_0 blank"></span>enc<span class="_0 blank"></span>iais L<span class="_0 blank"></span>in<span class="_0 blank"></span>ea<span class="_0 blank"></span>re<span class="_0 blank"></span>s Au<span class="_0 blank"></span>tôn<span class="_0 blank"></span>oma<span class="_0 blank"></span>s de P<span class="_0 blank"></span>rim<span class="_0 blank"></span>eir<span class="_0 blank"></span>a-O<span class="_0 blank"></span>rde<span class="_0 blank"></span>m</div><div class="t m0 x7 h14 y3a ff1 fs1 fc1 sc0 ls24 ws12">Além<span class="_5 blank"> </span>disso,<span class="_4 blank"> </span>sup ondo<span class="_5 blank"> </span><span class="ff2 ws3">y</span><span class="fs2 ls5d v7">0</span><span class="ff6 fs3 ls2b">6<span class="ff3 ls52">=</span><span class="ls5e">\ue000</span></span><span class="ff2 fs2 v5">b</span></div><div class="t m0 x1a h1e y3b ff2 fs2 fc1 sc0 ls5f">a<span class="ff1 fs1 ls24 ws2 v3">, a solução de uma equação autônoma</span></div><div class="t m0 x7 h4 y3c ff1 fs1 fc1 sc0 ls24 ws13">é tal que:</div><div class="t m0 x9 h1f y3d ff2 fs4 fc0 sc0 ls7">i<span class="ff4 ls8">.<span class="ff1 fc1 ls24 ws2b">Se <span class="ff2 ls60">a<span class="ff8 fs5 lsf"><</span></span><span class="ws2c">0, então:</span></span></span></div><div class="t m0 xf h1f y3e ff1 fs4 fc1 sc0 ls24">lim</div><div class="t m0 x0 h20 y3f ff2 fs2 fc1 sc0 ls61">t<span class="ff6 fs8 ls62">!<span class="ff9 fs9 ls24">\u221e</span></span></div><div class="t m0 x1b h8 y3e ff2 fs4 fc1 sc0 ls63">y<span class="ff3 fs5 ls15 v0">(</span><span class="lsb">t<span class="ff3 fs5 lsc v0">)<span class="ls64 v0">=</span></span><span class="ff1 ls24">lim</span></span></div><div class="t m0 x1c h21 y3f ff2 fs2 fc1 sc0 ls65">t<span class="ff6 fs8 ls66">!<span class="ff9 fs9 ls67">\u221e<span class="ff5 fs4 ls68 vd">\ue012</span></span></span><span class="fs4 ls24 ws2d ve">k<span class="_0 blank"></span>e <span class="fs2 ws16 v3">a t<span class="_5 blank"> </span></span><span class="ff6 fs5 ls69">\ue000</span><span class="v2">b</span></span></div><div class="t m0 x1d h22 y40 ff2 fs4 fc1 sc0 ls6a">a<span class="ff5 ls6b vf">\ue013</span><span class="ff3 fs5 ls6c v2">=<span class="ff6 ls6d">\ue000</span></span><span class="ls24 v6">b</span></div><div class="t m0 x1e h9 y40 ff2 fs4 fc1 sc0 ls24">a</div><div class="t m0 x1f h1f y41 ff1 fs4 fc1 sc0 ls24 wsc">Neste caso, a solução é <span class="fc2 v0">assintoticamente convergente <span class="fc1 ws2e">e<span class="_5 blank"> </span>o<span class="_5 blank"> </span>ponto</span></span></div><div class="t m0 x1f h23 y42 ff2 fs4 fc1 sc0 ls6e">y<span class="ff3 fs5 ls6f">=<span class="ff6 ls70">\ue000</span></span><span class="fs2 ls24 v3">b</span></div><div class="t m0 x19 h24 y43 ff2 fs2 fc1 sc0 ls71">a<span class="ff1 fs4 ls24 ws2f v10">é um <span class="fc2 wsc v0">equilíb<span class="_0 blank"></span>rio estável<span class="fc1">.</span></span></span></div><div class="t m0 xb h1f y44 ff2 fs4 fc0 sc0 ls24 wsf">ii <span class="ff4 ls13">.</span><span class="ff1 fc1 ws2b">Se <span class="ff2 ls60">a<span class="ff8 fs5 lsf">></span></span><span class="ws2c">0, então:</span></span></div><div class="t m0 xf h1f y45 ff1 fs4 fc1 sc0 ls24">lim</div><div class="t m0 x0 h20 y46 ff2 fs2 fc1 sc0 ls72">t<span class="ff6 fs8 ls66">!<span class="ff9 fs9 ls24">\u221e</span></span></div><div class="t m0 x1b h8 y45 ff2 fs4 fc1 sc0 ls73">y<span class="ff3 fs5 lsa v0">(</span><span class="ls74">t<span class="ff3 fs5 lsc v0">)<span class="ls64 v0">=</span></span><span class="ff1 ls24">lim</span></span></div><div class="t m0 xa h25 y46 ff2 fs2 fc1 sc0 ls65">t<span class="ff6 fs8 ls66">!<span class="ff9 fs9 ls67">\u221e<span class="ff5 fs4 ls75 vd">\ue012</span></span></span><span class="fs4 ls24 ws30 ve">k<span class="_0 blank"></span>e <span class="fs2 ws1e v3">a t<span class="_5 blank"> </span></span><span class="ff6 fs5 ls76">\ue000</span><span class="v2">b</span></span></div><div class="t m0 x1d h26 y47 ff2 fs4 fc1 sc0 ls77">a<span class="ff5 ls6b vf">\ue013</span><span class="ff3 fs5 ls6c v2">=<span class="ff6 ls78">\ue006<span class="ff9 fs4 ls24">\u221e</span></span></span></div><div class="t m0 x1f h1f y48 ff1 fs4 fc1 sc0 ls24 wsc">Neste caso, a solução é <span class="fc2 v0">assintoticamente divergente <span class="fc1 ws31">e<span class="_5 blank"> </span>que<span class="_5 blank"> </span>o<span class="_9 blank"> </span>p onto</span></span></div><div class="t m0 x1f h27 y49 ff2 fs4 fc1 sc0 ls6e">y<span class="ff3 fs5 ls6f">=<span class="ff6 ls70">\ue000</span></span><span class="fs2 ls24 v3">b</span></div><div class="t m0 x19 h28 y4a ff2 fs2 fc1 sc0 ls71">a<span class="ff1 fs4 ls24 ws2f v10">é um <span class="fc2 ws32 v0">equilíb<span class="_0 blank"></span>rio instável<span class="fc1">.</span></span></span></div><div class="t m0 x5 h5 y4b ff1 fs2 fc1 sc0 ls24">6</div></div></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div> <div id="pf3" class="pf w0 h0" data-page-no="3"><div class="pc pc3 w0 h0"><img fetchpriority="low" loading="lazy" class="bi x0 y0 w1 h1" alt="" src="https://files.passeidireto.com/ae95c9c7-2446-4b64-84fc-96c538ce870f/bg3.png"><div class="c x0 y1 w2 h2"><div class="t m0 x6 h3 y1f ff1 fs0 fc0 sc0 ls24 ws1a">Eq<span class="_0 blank"></span>.<span class="_1 blank"> </span>D<span class="_0 blank"></span>ifer<span class="_0 blank"></span>enc<span class="_0 blank"></span>iais L<span class="_0 blank"></span>in<span class="_0 blank"></span>ea<span class="_0 blank"></span>res A<span class="_0 blank"></span>ut<span class="_0 blank"></span>ôno<span class="_0 blank"></span>ma<span class="_0 blank"></span>s de P<span class="_0 blank"></span>rim<span class="_0 blank"></span>eir<span class="_0 blank"></span>a-Or<span class="_0 blank"></span>dem</div><div class="t m0 x7 h4 y4c ff1 fs1 fc2 sc0 ls24 ws33">Exemplo: <span class="fc1 ws20">Resolva a equação diferencial</span></div><div class="t m0 x20 h6 y4d ff2 fs1 fc1 sc0 ls24">dy</div><div class="t m0 x20 hf y4e ff2 fs1 fc1 sc0 ls24 ws8">dt <span class="ff3 fs3 ls57 v1">=<span class="ff6 ls79">\ue000</span></span><span class="ff1 ws3 v1">2</span><span class="ls7a v1">y<span class="ff3 fs3 ls7b">+</span></span><span class="ff1 ws3 v1">6<span class="ff4 ls7c">,</span><span class="ws18">com <span class="ff2 ls7d">y<span class="ff3 fs3 ls32 v0">(</span></span><span class="ls7e">0<span class="ff3 fs3 ls7f v0">)<span class="ls80 v0">=</span></span></span>10</span></span></div><div class="t m0 x7 h29 y4f ff1 fs1 fc1 sc0 ls24 ws2">Analise as p<span class="_0 blank"></span>ropriedades da trajetória da solução <span class="ff2 ls2d">y<span class="ff3 fs3 ls5 v0">(</span><span class="ls6 v0">t<span class="ff3 fs3 ls20 v0">)</span><span class="ff4 ls24">.</span></span></span></div><div class="t m0 x21 h2a y50 ffa fsa fc0 sc0 ls81">I<span class="ff1 fs4 fc1 ls24 ws34 v7">Re-escreva a equação acima como:</span></div><div class="t m0 xa h9 y51 ff2 fs4 fc1 sc0 ls24">dy</div><div class="t m0 xa ha y52 ff2 fs4 fc1 sc0 ls24 ws35">dt <span class="ff3 fs5 ls82 v2">+</span><span class="ff1 wsd v2">2</span><span class="ls6e v2">y<span class="ff3 fs5 lsf">=</span></span><span class="ff1 v2">6</span></div><div class="t m0 x1f h2b y53 ff1 fs4 fc1 sc0 ls24 ws36">e multiplique amb<span class="_3 blank"> </span>os os lados pelo fa<span class="_3 blank"> </span>to<span class="_0 blank"></span>r de integração <span class="ff2 ls83">e</span><span class="fs2 ls84 v10">2<span class="ff2 ls85">t</span></span><span class="ffb">:</span></div><div class="t m0 x4 h2c y54 ff2 fs4 fc1 sc0 ls86">e<span class="ff1 fs2 ls87 v3">2<span class="ff2 ls88">t</span></span><span class="ff5 ls89 v11">\ue012</span><span class="ls24 v2">dy</span></div><div class="t m0 x1c h2d y55 ff2 fs4 fc1 sc0 ls24 ws37">dt <span class="ff3 fs5 ls8a v2">+</span><span class="ff1 wsd v2">2</span><span class="ls8b v2">y</span><span class="ff5 ls8c vf">\ue013</span><span class="ff3 fs5 lsf v2">=</span><span class="ff1 wsd v2">6</span><span class="ls8d v2">e</span><span class="ff1 fs2 ls84 v12">2</span><span class="fs2 v12">t</span></div><div class="t m0 x5 h5 y56 ff1 fs2 fc1 sc0 ls24">7</div></div><div class="c x0 y7 w2 h2"><div class="t m0 x6 h3 y8 ff1 fs0 fc0 sc0 ls24 ws23">Eq<span class="_0 blank"></span>.<span class="_1 blank"> </span>D<span class="_0 blank"></span>ifer<span class="_0 blank"></span>enc<span class="_0 blank"></span>iais L<span class="_0 blank"></span>in<span class="_0 blank"></span>ea<span class="_0 blank"></span>re<span class="_0 blank"></span>s Au<span class="_0 blank"></span>tôn<span class="_0 blank"></span>oma<span class="_0 blank"></span>s de P<span class="_0 blank"></span>rim<span class="_0 blank"></span>eir<span class="_0 blank"></span>a-O<span class="_0 blank"></span>rde<span class="_0 blank"></span>m</div><div class="t m0 x7 h4 y57 ff1 fs1 fc1 sc0 ls24 ws3">(Cont.)</div><div class="t m0 x21 h2a y58 ffa fsa fc0 sc0 ls81">I<span class="ff1 fs4 fc1 ls24 ws36 v7">Note que, integrando amb<span class="_3 blank"> </span>os os lados em relação a <span class="ff2 ls8e">t</span><span class="wsc">, obtemos:</span></span></div><div class="t m0 x22 h2e y59 ff5 fsb fc1 sc0 ls8f">Z<span class="ff2 fs4 ls90 v13">e</span><span class="ff1 fs2 ls87 v14">2<span class="ff2 ls88">t</span></span><span class="fs4 ls91 v15">\ue012<span class="ff2 ls24 wsd v16">dy</span></span></div><div class="t m0 x0 h2f y5a ff2 fs4 fc1 sc0 ls24 ws38">dt <span class="ff3 fs5 ls82 v2">+</span><span class="ff1 wsd v2">2</span><span class="ls92 v2">y</span><span class="ff5 ls93 vf">\ue013</span><span class="ws39 v2">dt <span class="ff3 fs5 lsd">=<span class="ff1 fs4 ls94">6<span class="ff5 fsb ls8f vb">Z</span></span></span></span><span class="ls86 v2">e</span><span class="ff1 fs2 ls84 v12">2<span class="ff2 ls95">t</span></span><span class="ws3a v2">dt <span class="ff6 fs5 ls96">)</span><span class="ls8d">e<span class="ff1 fs2 ls84 v3">2<span class="ff2 ls97">t</span></span><span class="ls98">y<span class="ff3 fs5 lsd">=</span><span class="ls99">k<span class="ff3 fs5 ls8a">+</span><span class="ff1 ls24 wsd">3</span><span class="ls83">e<span class="ff1 fs2 ls87 v3">2</span></span></span></span></span></span><span class="fs2 v12">t</span></div><div class="t m0 x1f h30 y5b ff1 fs4 fc1 sc0 ls24 wsc">P<span class="_0 blank"></span>ortanto, a <span class="fc2 v0">solução geral <span class="fc1 ws3b">do<span class="_9 blank"> </span>problema<span class="_9 blank"> </span>é<span class="_5 blank"> </span>dada<span class="_9 blank"> </span>p or:</span></span></div><div class="t m0 x1a h2b y5c ff2 fs4 fc1 sc0 ls73">y<span class="ff3 fs5 lsa v0">(</span><span class="ls74">t<span class="ff3 fs5 lsc v0">)<span class="lsf v0">=</span></span><span class="ls24 ws3c">k<span class="_0 blank"></span>e <span class="ff6 fs8 ls9a v10">\ue000<span class="ff1 fs2 ls9b">2<span class="ff2 ls9c">t</span></span></span><span class="ff3 fs5 ls8a">+</span><span class="ff1">3</span></span></span></div><div class="t m0 x21 h31 y5d ffa fsa fc0 sc0 ls81">I<span class="ff1 fs4 fc1 ls24 ws3d v7">Dada a condição inicial <span class="ff2 ls63">y<span class="ff3 fs5 ls15 v0">(</span><span class="ff1 ls9d">0<span class="ff3 fs5 lsc v0">)<span class="lsf v0">=</span></span><span class="ls24 ws3e">10, então devemos ter que:</span></span></span></span></div><div class="t m0 x23 h8 y5e ff2 fs4 fc1 sc0 ls73">y<span class="ff3 fs5 ls9e v0">(</span><span class="ff1 ls9f">0<span class="ff3 fs5 lsc v0">)<span class="lsf v0">=</span></span></span><span class="lsa0">k<span class="ff3 fs5 ls8a">+</span><span class="ff1 lsa1">3<span class="ff3 fs5 lsd">=</span><span class="ls24 ws3f">10 <span class="ff6 fs5 ls96">)</span></span></span><span class="lsa2">k<span class="ff3 fs5 lsd">=</span><span class="ff1 ls24">7</span></span></span></div><div class="t m0 x1f h1f y5f ff1 fs4 fc1 sc0 ls24 ws2c">Logo, a <span class="fc2 ws40">solução pa<span class="_0 blank"></span>rticular <span class="fc1 ws2e">é<span class="_9 blank"> </span>dada<span class="_5 blank"> </span>p or:</span></span></div><div class="t m0 x1a h2b y60 ff2 fs4 fc1 sc0 ls73">y<span class="ff3 fs5 lsa3 v0">(</span><span class="ls74">t<span class="ff3 fs5 lsa4 v0">)<span class="lsd v0">=</span></span><span class="ff1 ls24 wsd">7</span><span class="lsa5">e<span class="ff6 fs8 lsa6 v10">\ue000<span class="ff1 fs2 ls87">2<span class="ff2 lsa7">t</span></span></span><span class="ff3 fs5 ls82">+</span><span class="ff1 ls24">3</span></span></span></div><div class="t m0 x5 h5 y39 ff1 fs2 fc1 sc0 ls24">8</div></div><div class="c x0 y15 w2 h2"><div class="t m0 x6 h3 y16 ff1 fs0 fc0 sc0 ls24 ws23">Eq<span class="_0 blank"></span>.<span class="_1 blank"> </span>D<span class="_0 blank"></span>ifer<span class="_0 blank"></span>enc<span class="_0 blank"></span>iais L<span class="_0 blank"></span>in<span class="_0 blank"></span>ea<span class="_0 blank"></span>re<span class="_0 blank"></span>s Au<span class="_0 blank"></span>tôn<span class="_0 blank"></span>oma<span class="_0 blank"></span>s de P<span class="_0 blank"></span>rim<span class="_0 blank"></span>eir<span class="_0 blank"></span>a-O<span class="_0 blank"></span>rde<span class="_0 blank"></span>m</div><div class="t m0 x7 h4 y61 ff1 fs1 fc1 sc0 ls24 ws3">(Cont.)</div><div class="t m0 x21 h32 y62 ffa fsa fc0 sc0 ls81">I<span class="ff1 fs4 fc1 ls24 wsc v7">Note que o <span class="fc2 ws41">p onto<span class="_9 blank"> </span>de<span class="_5 blank"> </span>equilíbrio<span class="_9 blank"> </span><span class="fc1 ws42">desta equação é dado p<span class="_3 blank"> </span>o<span class="_0 blank"></span>r:</span></span></span></div><div class="t m0 x24 h9 y63 ff2 fs4 fc1 sc0 ls24">dy</div><div class="t m0 x24 h33 y64 ff2 fs4 fc1 sc0 ls24 ws43">dt <span class="ff3 fs5 lsf v2">=</span><span class="ff1 lsa8 v2">0</span><span class="ff6 fs5 ws44 v2">$<span class="_a blank"> </span>\ue000 </span><span class="ff1 wsd v2">2</span><span class="lsa9 v2">y<span class="ff3 fs5 lsaa">+<span class="ff1 fs4 lsab">6</span><span class="lsd">=<span class="ff1 fs4 lsac">0</span><span class="ff6 lsad">)</span></span></span></span><span class="ls6e v2">y<span class="ff3 fs5 lsf">=</span></span><span class="ff1 v2">3</span></div><div class="t m0 x21 h34 y65 ffa fsa fc0 sc0 ls81">I<span class="ff1 fs4 fc1 ls24 ws2c v7">Além disso, como</span></div><div class="t m0 x3 h1f y66 ff1 fs4 fc1 sc0 ls24">lim</div><div class="t m0 x25 h20 y67 ff2 fs2 fc1 sc0 ls72">t<span class="ff6 fs8 ls66">!<span class="ff9 fs9 ls24">\u221e</span></span></div><div class="t m0 x26 h8 y66 ff2 fs4 fc1 sc0 ls63">y<span class="ff3 fs5 ls15 v0">(</span><span class="lsae">t<span class="ff3 fs5 lsc v0">)<span class="lsaf v0">=</span></span><span class="ff1 ls24">lim</span></span></div><div class="t m0 x27 h35 y67 ff2 fs2 fc1 sc0 ls61">t<span class="ff6 fs8 lsb0">!<span class="ff9 fs9 lsb1">\u221e<span class="ff5 fs4 lsb2 v17">\ue010<span class="ff1 lsb3 v18">7<span class="ff2 lsb4">e</span></span></span></span><span class="ls9a v19">\ue000</span></span><span class="ff1 ls87 v19">2</span><span class="lsb5 v19">t</span><span class="ff3 fs5 ls8a ve">+<span class="ff1 fs4 lsb6">3<span class="ff5 lsb7 v12">\ue011</span></span><span class="lsb8">=<span class="ff1 fs4 ls24 wsd">3<span class="ff4">,</span></span></span></span></div><div class="t m0 x1f h36 y68 ff1 fs4 fc1 sc0 ls24 wsc">então a solução é <span class="fc2 v0">assintoticamente convergente <span class="fc1 ws2e">e<span class="_5 blank"> </span>o<span class="_9 blank"> </span>p onto<span class="_5 blank"> </span><span class="ff2 ls6e">y<span class="ff3 fs5 lsf">=</span></span><span class="wsc">3 é um</span></span></span></div><div class="t m0 x1f h1f y69 ff1 fs4 fc2 sc0 ls24 wsc">equilíb<span class="_0 blank"></span>rio estável<span class="fc1">.</span></div><div class="t m0 x5 h5 y1e ff1 fs2 fc1 sc0 ls24">9</div></div></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div> <div id="pf4" class="pf w0 h0" data-page-no="4"><div class="pc pc4 w0 h0"><img fetchpriority="low" loading="lazy" class="bi x0 y0 w1 h1" alt="" src="https://files.passeidireto.com/ae95c9c7-2446-4b64-84fc-96c538ce870f/bg4.png"><div class="c x0 y1 w2 h2"><div class="t m0 x10 h37 y6a ff1 fsc fc1 sc0 ls24 ws45">D<span class="_b blank"></span>i<span class="_0 blank"></span>a<span class="_0 blank"></span>g<span class="_0 blank"></span>ra<span class="_b blank"></span>m<span class="_b blank"></span>a d<span class="_b blank"></span>e F<span class="_b blank"></span>a<span class="_b blank"></span>se<span class="_b blank"></span>s</div><div class="t m0 x28 h5 y6b ff1 fs2 fc1 sc0 ls24 ws46">10</div></div><div class="c x0 y7 w2 h2"><div class="t m0 x6 h3 y8 ff1 fs0 fc0 sc0 ls24 ws0">D<span class="_0 blank"></span>iagr<span class="_0 blank"></span>am<span class="_0 blank"></span>a de F<span class="_b blank"></span>ases - S<span class="_0 blank"></span>olu<span class="_0 blank"></span>ção Qu<span class="_0 blank"></span>al<span class="_0 blank"></span>itat<span class="_0 blank"></span>iva</div><div class="t m0 x7 h4 y6c ff1 fs1 fc1 sc0 ls24 ws7">Considere uma equação diferencial de p<span class="_0 blank"></span>rimeira ordem <span class="fc2 ws1b">autônoma </span>do</div><div class="t m0 x7 h4 y6d ff1 fs1 fc1 sc0 ls24 ws2a">tip o:</div><div class="t m0 x29 h6 y6e ff2 fs1 fc1 sc0 ls24">dy</div><div class="t m0 x29 h7 y6f ff2 fs1 fc1 sc0 ls24 ws8">dt <span class="ff3 fs3 ls25 v1">=</span><span class="lsb9 v1">f</span><span class="ff3 fs3 ls18 v1">(</span><span class="lsba v1">y</span><span class="ff3 fs3 v1">)</span></div><div class="t m0 x7 h38 y70 ff1 fs1 fc1 sc0 ls24 ws2">Neste caso, será difícil encontra<span class="_0 blank"></span>r uma solução explícita para <span class="ff2 ls2d">y<span class="ff3 fs3 ls5 v0">(</span><span class="ls6">t<span class="ff3 fs3 ls18 v0">)</span></span></span>.</div><div class="t m0 x7 h4 y71 ff1 fs1 fc1 sc0 ls24 ws2">Ainda assim, p<span class="_3 blank"> </span>o<span class="_3 blank"> </span>demos obter algumas p<span class="_0 blank"></span>ropriedades importantes da</div><div class="t m0 x7 h4 y72 ff1 fs1 fc1 sc0 ls24 ws47">solução através de um grá\u2026co chamado de <span class="fc2 ws48 v0">diagrama de fases <span class="fc1 ws3">(solução</span></span></div><div class="t m0 x7 h4 y73 ff1 fs1 fc1 sc0 ls24 ws3">qualitativa).</div><div class="t m0 x28 h5 y74 ff1 fs2 fc1 sc0 ls24 ws46">11</div></div><div class="c x0 y15 w2 h2"><div class="t m0 x6 h3 y16 ff1 fs0 fc0 sc0 ls24 ws0">D<span class="_0 blank"></span>iagr<span class="_0 blank"></span>am<span class="_0 blank"></span>a de F<span class="_b blank"></span>ases - S<span class="_0 blank"></span>olu<span class="_0 blank"></span>ção Qu<span class="_0 blank"></span>al<span class="_0 blank"></span>itat<span class="_0 blank"></span>iva</div><div class="t m0 x7 h39 y75 ff1 fs1 fc1 sc0 lsbb">O<span class="fc2 ls24 ws49">diagrama de fases </span><span class="ls24 ws2">é construído traçando o grá\u2026co de <span class="ff2 lsb9">f<span class="ff3 fs3 ls32 v0">(</span><span class="lsbc">y<span class="ff3 fs3 ls18 v0">)</span></span></span>:</span></div><div class="t m0 x7 h4 y76 ff1 fs1 fc1 sc0 ls24 ws5">O<span class="_5 blank"> </span>p onto<span class="_4 blank"> </span><span class="ff2 lsbd">y</span><span class="ws4a">é um <span class="fc2 ws2a">p onto<span class="_5 blank"> </span>de<span class="_4 blank"> </span>equilíb<span class="_0 blank"></span>rio<span class="fc1 ws1">.<span class="_2 blank"> </span>De forma geral, uma equação</span></span></span></div><div class="t m0 x7 h4 y77 ff1 fs1 fc1 sc0 ls24 ws4b">diferencial p<span class="_3 blank"> </span>o<span class="_3 blank"> </span>de ter um equilíb<span class="_0 blank"></span>rio, mais de um ou nenhum equilíb<span class="_0 blank"></span>rio.</div><div class="t m0 x28 h5 y1e ff1 fs2 fc1 sc0 ls24 ws46">12</div></div></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div> <div id="pf5" class="pf w0 h0" data-page-no="5"><div class="pc pc5 w0 h0"><img fetchpriority="low" loading="lazy" class="bi x0 y0 w1 h1" alt="" src="https://files.passeidireto.com/ae95c9c7-2446-4b64-84fc-96c538ce870f/bg5.png"><div class="c x0 y1 w2 h2"><div class="t m0 x6 h3 y1f ff1 fs0 fc0 sc0 ls24 ws4c">D<span class="_0 blank"></span>iagr<span class="_0 blank"></span>am<span class="_0 blank"></span>a de F<span class="_b blank"></span>ases - S<span class="_0 blank"></span>olu<span class="_0 blank"></span>ção Qu<span class="_0 blank"></span>al<span class="_0 blank"></span>itat<span class="_0 blank"></span>iva</div><div class="t m0 x7 h4 y78 ff1 fs1 fc2 sc0 ls24 ws13">Equilíb<span class="_0 blank"></span>rio instável:</div><div class="t m0 x7 h3a y79 ff1 fs1 fc1 sc0 ls24 ws13">Neste caso, a função <span class="ff2 lsbe">f<span class="ff3 fs3 ls5 v0">(</span><span class="lsba">y<span class="ff3 fs3 lsbf v0">)</span></span></span><span class="ws2">cruza o eixo-<span class="ff2 lsc0">x</span>de baixo pa<span class="_0 blank"></span>ra cima.</span></div><div class="t m0 x28 h5 y7a ff1 fs2 fc1 sc0 ls24 ws46">13</div></div><div class="c x0 y7 w2 h2"><div class="t m0 x6 h3 y8 ff1 fs0 fc0 sc0 ls24 ws0">D<span class="_0 blank"></span>iagr<span class="_0 blank"></span>am<span class="_0 blank"></span>a de F<span class="_b blank"></span>ases - S<span class="_0 blank"></span>olu<span class="_0 blank"></span>ção Qu<span class="_0 blank"></span>al<span class="_0 blank"></span>itat<span class="_0 blank"></span>iva</div><div class="t m0 x7 h4 y7b ff1 fs1 fc2 sc0 ls24 ws13">Equilíb<span class="_0 blank"></span>rio estável:</div><div class="t m0 x7 h3b y7c ff1 fs1 fc1 sc0 ls24 ws13">Neste caso, a função <span class="ff2 lsbe">f<span class="ff3 fs3 ls5 v0">(</span><span class="lsba">y<span class="ff3 fs3 lsbf v0">)</span></span></span><span class="ws2">cruza o eixo-<span class="ff2 lsc0">x</span><span class="ws1">de cima para baixo.</span></span></div><div class="t m0 x28 h5 y7d ff1 fs2 fc1 sc0 ls24 ws46">14</div></div><div class="c x0 y15 w2 h2"><div class="t m0 x6 h3 y16 ff1 fs0 fc0 sc0 ls24 ws4d">Ex<span class="_0 blank"></span>em<span class="_0 blank"></span>plo</div><div class="t m0 x7 h3a y7e ff1 fs1 fc1 sc0 ls24 ws21">Considere a va<span class="_0 blank"></span>riável <span class="ff2 lsc1">y<span class="ff3 fs3 ls32 v0">(</span><span class="ls6">t<span class="ff3 fs3 lsbf v0">)</span></span></span><span class="ws7">cuja trajetória temp<span class="_3 blank"> </span>oral é governada pela</span></div><div class="t m0 x7 h4 y7f ff1 fs1 fc1 sc0 ls24 ws21">seguinte equação diferencial logística:</div><div class="t m0 xf h6 y80 ff2 fs1 fc1 sc0 ls24">dy</div><div class="t m0 xf hf y81 ff2 fs1 fc1 sc0 ls24 ws8">dt <span class="ff3 fs3 ls57 v1">=<span class="ls18 v0">(</span></span><span class="lsc2 v1">a<span class="ff6 fs3 ls3">\ue000</span></span><span class="ws4e v1">b<span class="_0 blank"></span>y <span class="ff3 fs3 ls20 v0">)</span><span class="lsc3">y<span class="ff4 ls29">,</span><span class="lsc4">a<span class="ff4 lsc5">,</span><span class="lsc6">b<span class="ff8 fs3 ls25">></span><span class="ff1 ls24">0</span></span></span></span></span></div><div class="t m0 x7 h4 y82 ff1 fs1 fc1 sc0 ls24 ws1">Encontre o(s) valo<span class="_0 blank"></span>res de equilíbrio de <span class="ff2 lsc7">y</span><span class="ws4f">e veri\u2026que se são estáveis ou</span></div><div class="t m0 x7 h4 y83 ff1 fs1 fc1 sc0 ls24 ws3">instáveis.</div><div class="t m0 x21 h32 y84 ffa fsa fc0 sc0 ls81">I<span class="ff1 fs4 fc1 ls24 ws50 v7">O diagrama de fases da equação diferencial acima é o seguinte:</span></div><div class="t m0 x28 h5 y1e ff1 fs2 fc1 sc0 ls24 ws46">15</div></div></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div> <div id="pf6" class="pf w0 h0" data-page-no="6"><div class="pc pc6 w0 h0"><img fetchpriority="low" loading="lazy" class="bi x0 y0 w1 h1" alt="" src="https://files.passeidireto.com/ae95c9c7-2446-4b64-84fc-96c538ce870f/bg6.png"><div class="c x0 y1 w2 h2"><div class="t m0 x6 h3 y1f ff1 fs0 fc0 sc0 ls24 ws4d">Ex<span class="_0 blank"></span>em<span class="_0 blank"></span>plo</div><div class="t m0 x7 h4 y85 ff1 fs1 fc1 sc0 ls24 ws3">(Cont.)</div><div class="t m0 x21 h32 y86 ffa fsa fc0 sc0 ls81">I<span class="ff1 fs4 fc1 ls24 wsc v7">Note que a equação p<span class="_3 blank"> </span>ossui dois equilíb<span class="_0 blank"></span>rios:</span></div><div class="t m0 x2a h3c y87 ff2 fsb fc0 sc0 lsc8">i<span class="ffc lsc9">.</span><span class="fc1 lsca">y<span class="ff1 fsd lscb v1a">1</span><span class="ff3 fse lscc v0">=</span><span class="ff1 ls24 ws51">0<span class="_9 blank"> </span>é<span class="_9 blank"> </span>u m<span class="_5 blank"> </span>eq u il íbrio<span class="_5 blank"> </span>inst á v e l;<span class="_5 blank"> </span>e</span></span></div><div class="t m0 x2b h3d y88 ff2 fsb fc0 sc0 ls24 ws52">ii <span class="ffc lscd">.</span><span class="fc1 lsca">y<span class="ff1 fsd lscb v1a">2</span><span class="ff3 fse lsce v0">=</span><span class="fsd ls24 v10">a</span></span></div><div class="t m0 x2c h3e y89 ff2 fsd fc1 sc0 lscf">b<span class="ff1 fsb ls24 ws53 v1b">é<span class="_9 blank"> </span>um<span class="_5 blank"> </span>eq u il íbrio<span class="_5 blank"> </span>est á v e l.</span></div><div class="t m0 x28 h5 y6b ff1 fs2 fc1 sc0 ls24 ws46">16</div></div><div class="c x0 y7 w2 h2"><div class="t m0 x2d h37 y8a ff1 fsc fc1 sc0 ls24 ws54">E<span class="_b blank"></span>q<span class="_0 blank"></span>u<span class="_b blank"></span>a<span class="_0 blank"></span>çõ<span class="_b blank"></span>e<span class="_0 blank"></span>s S<span class="_b blank"></span>ep<span class="_b blank"></span>a<span class="_b blank"></span>r<span class="_b blank"></span>á<span class="_0 blank"></span>v<span class="_0 blank"></span>e<span class="_0 blank"></span>is d<span class="_b blank"></span>e P<span class="_b blank"></span>r<span class="_0 blank"></span>im<span class="_b blank"></span>e<span class="_b blank"></span>ir<span class="_b blank"></span>a O<span class="_b blank"></span>rd<span class="_b blank"></span>e<span class="_b blank"></span>m</div><div class="t m0 x28 h5 y39 ff1 fs2 fc1 sc0 ls24 ws46">17</div></div><div class="c x0 y15 w2 h2"><div class="t m0 x6 h3 y16 ff1 fs0 fc0 sc0 ls24 ws4">Eq<span class="_0 blank"></span>ua<span class="_0 blank"></span>ções S<span class="_0 blank"></span>epa<span class="_b blank"></span>rá<span class="_0 blank"></span>vei<span class="_0 blank"></span>s de Pr<span class="_0 blank"></span>im<span class="_0 blank"></span>eira<span class="_0 blank"></span>-Or<span class="_0 blank"></span>dem</div><div class="t m0 x7 h4 y8b ff1 fs1 fc1 sc0 ls24 ws55">As <span class="fc2 ws56">equações diferenciais sepa<span class="_0 blank"></span>ráveis <span class="fc1 ws12">p ossuem<span class="_4 blank"> </span>a<span class="_5 blank"> </span>seguinte<span class="_4 blank"> </span>fo<span class="_0 blank"></span>rma:</span></span></div><div class="t m0 x2e h6 y8c ff2 fs1 fc1 sc0 ls24">dy</div><div class="t m0 x2e hf y8d ff2 fs1 fc1 sc0 ls24 ws8">dt <span class="ff3 fs3 ls25 v1">=</span><span class="lsd0 v1">g</span><span class="ff3 fs3 ls32 v1">(</span><span class="lsbc v1">y</span><span class="ff3 fs3 lsd1 v1">)</span><span class="lsd2 v1">h</span><span class="ff3 fs3 ls18 v1">(</span><span class="ls6 v1">t</span><span class="ff3 fs3 ls20 v1">)</span><span class="ff4 v1">,</span></div><div class="t m0 x7 h4 y8e ff1 fs1 fc1 sc0 ls24 ws21">i.e.<span class="_2 blank"> </span>o lado direito da equação p<span class="_3 blank"> </span>o<span class="_3 blank"> </span>de ser escrito como o p<span class="_0 blank"></span>ro<span class="_3 blank"> </span>duto de um</div><div class="t m0 x7 h3f y8f ff1 fs1 fc1 sc0 ls24 ws4f">fato<span class="_0 blank"></span>r que dep<span class="_3 blank"> </span>ende somente de <span class="ff2 lsc3">y<span class="ff4 lsd3">,</span><span class="fc2 lsd4 v0">g<span class="ff3 fs3 ls18">(</span><span class="lsbc">y<span class="ff3 fs3 ls18">)</span><span class="ff4 fc1 lsd5">,</span></span></span></span><span class="ws1 v0">e de um outro fator que</span></div><div class="t m0 x7 h3f y90 ff1 fs1 fc1 sc0 ls24 ws2a">dep ende<span class="_5 blank"> </span>somente<span class="_4 blank"> </span>de<span class="_5 blank"> </span><span class="ff2 lsd6">t<span class="ff4 lsd7">,</span><span class="fc2 lsd8 v0">h<span class="ff3 fs3 ls32">(</span><span class="ls6">t<span class="ff3 fs3 ls5">)</span></span></span></span><span class="ff4 v0">.</span></div><div class="t m0 x28 h5 y91 ff1 fs2 fc1 sc0 ls24 ws46">18</div></div></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div> <div id="pf7" class="pf w0 h0" data-page-no="7"><div class="pc pc7 w0 h0"><img fetchpriority="low" loading="lazy" class="bi x0 y0 w1 h1" alt="" src="https://files.passeidireto.com/ae95c9c7-2446-4b64-84fc-96c538ce870f/bg7.png"><div class="c x0 y1 w2 h2"><div class="t m0 x6 h3 y1f ff1 fs0 fc0 sc0 ls24 ws57">Eq<span class="_0 blank"></span>ua<span class="_0 blank"></span>ções S<span class="_0 blank"></span>epa<span class="_b blank"></span>rá<span class="_0 blank"></span>vei<span class="_0 blank"></span>s de Pr<span class="_0 blank"></span>im<span class="_0 blank"></span>eira<span class="_0 blank"></span>-Or<span class="_0 blank"></span>dem</div><div class="t m0 x7 h4 y92 ff1 fs1 fc2 sc0 ls24 ws21">Exemplo 1:<span class="_2 blank"> </span><span class="fc1 ws1">As seguintes equações diferenciais são sepa<span class="_0 blank"></span>ráveis:</span></div><div class="t m0 x2e h6 y93 ff2 fs1 fc1 sc0 ls24">dy</div><div class="t m0 x4 h40 y94 ff2 fs1 fc1 sc0 ls24 ws8">dt <span class="ff3 fs3 ls25 v1">=</span><span class="lsd9 v1">y</span><span class="ff1 fs2 lsda v9">2</span><span class="ff5 lsdb v1c">\ue000</span><span class="lsdc v1">t</span><span class="ff1 fs2 lsdd v9">2</span><span class="ff3 fs3 ls7b v1">+</span><span class="lsde v1">t</span><span class="ff5 v1c">\ue001</span></div><div class="t m0 x2f h6 y95 ff2 fs1 fc1 sc0 ls24">dy</div><div class="t m0 x2f h41 y96 ff2 fs1 fc1 sc0 ls24 ws8">dt <span class="ff3 fs3 ls25 v1">=</span><span class="ls37 v1">e</span><span class="fs2 lsdf v9">y</span><span class="ls37 v1">e</span><span class="fs2 v9">t</span></div><div class="t m0 x30 h6 y97 ff2 fs1 fc1 sc0 ls24">dy</div><div class="t m0 x30 h42 y98 ff2 fs1 fc1 sc0 ls24 ws58">dt <span class="ff3 fs3 ls57 v1">=<span class="ls18 v0">(</span></span><span class="lse0 v1">y<span class="ff3 fs3 ls7b">+<span class="ff1 fs1 lse1">1</span><span class="lse2 v0">)</span></span></span><span class="ff1 v1d">1</span></div><div class="t m0 x31 h6 y98 ff2 fs1 fc1 sc0 ls24">t</div><div class="t m0 x32 h6 y99 ff2 fs1 fc1 sc0 ls24">dy</div><div class="t m0 x32 h43 y9a ff2 fs1 fc1 sc0 ls24 ws8">dt <span class="ff3 fs3 ls25 v1">=</span><span class="lsc3 v1">y</span><span class="ff1 fs2 lse3 v9">2</span><span class="ff3 fs3 ls3 v1">+</span><span class="ff1 v1">1</span></div><div class="t m0 x28 h5 y6 ff1 fs2 fc1 sc0 ls24 ws46">19</div></div><div class="c x0 y7 w2 h2"><div class="t m0 x6 h3 y8 ff1 fs0 fc0 sc0 ls24 ws4">Eq<span class="_0 blank"></span>ua<span class="_0 blank"></span>ções S<span class="_0 blank"></span>epa<span class="_b blank"></span>rá<span class="_0 blank"></span>vei<span class="_0 blank"></span>s de Pr<span class="_0 blank"></span>im<span class="_0 blank"></span>eira<span class="_0 blank"></span>-Or<span class="_0 blank"></span>dem</div><div class="t m0 x7 h4 y9b ff1 fs1 fc2 sc0 ls24 ws21">Exemplo 2:<span class="_2 blank"> </span><span class="fc1 ws1">As seguintes equações diferenciais são não-sepa<span class="_0 blank"></span>ráveis:</span></div><div class="t m0 x33 h6 y9c ff2 fs1 fc1 sc0 ls24">dy</div><div class="t m0 x33 h44 y9d ff2 fs1 fc1 sc0 ls24 ws8">dt <span class="ff3 fs3 ls25 v1">=</span><span class="lsd9 v1">y</span><span class="ff1 fs2 lse4 v9">2</span><span class="ff3 fs3 ls3 v1">+</span><span class="lsd6 v1">t</span><span class="ff1 fs2 v9">2</span></div><div class="t m0 x26 h6 y9e ff2 fs1 fc1 sc0 ls24">dy</div><div class="t m0 x26 h7 y9f ff2 fs1 fc1 sc0 ls24 ws58">dt <span class="ff3 fs3 ls25 v1">=</span><span class="lse5 v1">a</span><span class="ff3 fs3 ls18 v1">(</span><span class="ls33 v1">t</span><span class="ff3 fs3 ls20 v1">)</span><span class="ls2 v1">y<span class="ff3 fs3 ls27">+</span><span class="lse6">b<span class="ff3 fs3 ls18 v0">(</span><span class="ls33">t</span></span></span><span class="ff3 fs3 v1">)</span></div><div class="t m0 x30 h6 ya0 ff2 fs1 fc1 sc0 ls24">dy</div><div class="t m0 x30 h45 ya1 ff2 fs1 fc1 sc0 ls24 ws8">dt <span class="ff3 fs3 ls0 v1">=</span><span class="ws59 v1">t<span class="_0 blank"></span>y <span class="ff3 fs3 ls3">+</span><span class="lse7">t<span class="ff1 fs2 lse8 v5">2</span><span class="lsd9">y<span class="ff1 fs2 ls24 v5">2</span></span></span></span></div><div class="t m0 x28 h5 ya2 ff1 fs2 fc1 sc0 ls24 ws46">20</div></div><div class="c x0 y15 w2 h2"><div class="t m0 x6 h3 y16 ff1 fs0 fc0 sc0 ls24 ws4">Eq<span class="_0 blank"></span>ua<span class="_0 blank"></span>ções S<span class="_0 blank"></span>epa<span class="_b blank"></span>rá<span class="_0 blank"></span>vei<span class="_0 blank"></span>s de Pr<span class="_0 blank"></span>im<span class="_0 blank"></span>eira<span class="_0 blank"></span>-Or<span class="_0 blank"></span>dem</div><div class="t m0 x7 h4 ya3 ff1 fs1 fc1 sc0 ls24 ws1">Dada uma equação sepa<span class="_0 blank"></span>rável:</div><div class="t m0 x2e h6 ya4 ff2 fs1 fc1 sc0 ls24">dy</div><div class="t m0 x2e hf ya5 ff2 fs1 fc1 sc0 ls24 ws8">dt <span class="ff3 fs3 ls25 v1">=</span><span class="lsd0 v1">g</span><span class="ff3 fs3 ls32 v1">(</span><span class="lsbc v1">y</span><span class="ff3 fs3 lsd1 v1">)</span><span class="lsd2 v1">h</span><span class="ff3 fs3 ls18 v1">(</span><span class="ls6 v1">t</span><span class="ff3 fs3 ls20 v1">)</span><span class="ff4 v1">,</span></div><div class="t m0 x7 h4 ya6 ff1 fs1 fc1 sc0 ls24 ws1">re-escreva-a como:<span class="_c blank"> </span><span class="v14">1</span></div><div class="t m0 x34 h7 ya7 ff2 fs1 fc1 sc0 lse9">g<span class="ff3 fs3 ls18 v0">(</span><span class="lsbc">y<span class="ff3 fs3 lsea v0">)</span><span class="ls24 ws5a v1">dy <span class="ff3 fs3 ls0">=</span><span class="lsd2">h<span class="ff3 fs3 ls18 v0">(</span><span class="ls33">t<span class="ff3 fs3 ls20 v0">)</span><span class="ls24">dt</span></span></span></span></span></div><div class="t m0 x7 h4 ya8 ff1 fs1 fc1 sc0 ls24 ws5b">Integrando o lado esquerdo em relação a <span class="ff2 lseb">y</span><span class="ws5c">e o lado direito em relação</span></div><div class="t m0 x7 h4 ya9 ff1 fs1 fc1 sc0 lsec">a<span class="ff2 ls5a">t</span><span class="ls24 ws21">, obtemos:<span class="_d blank"> </span><span class="ff5 fs7 lsed v1e">Z</span><span class="v1f">1</span></span></div><div class="t m0 x35 h46 yaa ff2 fs1 fc1 sc0 lse9">g<span class="ff3 fs3 ls32 v0">(</span><span class="lsbc">y<span class="ff3 fs3 lsee v0">)</span><span class="ls24 ws5d v1">dy <span class="ff3 fs3 ls57">=<span class="ff5 fs7 ls2e v6">Z</span></span></span><span class="lsd2 v1">h</span><span class="ff3 fs3 ls18 v1">(</span><span class="ls33 v1">t</span><span class="ff3 fs3 ls20 v1">)</span><span class="ls24 ws5e v1">dt <span class="ff3 fs3 ls3">+</span>k</span></span></div><div class="t m0 x7 h47 yab ff1 fs1 fc1 sc0 ls24 ws5f">Resolva esta equação pa<span class="_0 blank"></span>ra <span class="ff2 ls31">y<span class="ff3 fs3 ls18 v0">(</span><span class="ls6">t<span class="ff3 fs3 ls20 v0">)</span></span></span><span class="ff4">.</span></div><div class="t m0 x28 h5 y1e ff1 fs2 fc1 sc0 ls24 ws46">21</div></div></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div> <div id="pf8" class="pf w0 h0" data-page-no="8"><div class="pc pc8 w0 h0"><img fetchpriority="low" loading="lazy" class="bi x0 yac w1 h48" alt="" src="https://files.passeidireto.com/ae95c9c7-2446-4b64-84fc-96c538ce870f/bg8.png"><div class="c x0 y1 w2 h2"><div class="t m0 x6 h3 y1f ff1 fs0 fc0 sc0 ls24 ws57">Eq<span class="_0 blank"></span>ua<span class="_0 blank"></span>ções S<span class="_0 blank"></span>epa<span class="_b blank"></span>rá<span class="_0 blank"></span>vei<span class="_0 blank"></span>s de Pr<span class="_0 blank"></span>im<span class="_0 blank"></span>eira<span class="_0 blank"></span>-Or<span class="_0 blank"></span>dem</div><div class="t m0 x7 h4 yad ff1 fs1 fc2 sc0 ls24 ws21">Exemplo 1:<span class="_2 blank"> </span><span class="fc1">Considere a seguinte equação:</span></div><div class="t m0 x2e h6 yae ff2 fs1 fc1 sc0 ls24">dy</div><div class="t m0 x2e h49 yaf ff2 fs1 fc1 sc0 ls24 ws8">dt <span class="ff3 fs3 ls25 v1">=</span><span class="ff1 lsef v1">3</span><span class="ff3 fs3 ls32 v1">(</span><span class="ff1 lsf0 v1">2<span class="ff6 fs3 ls3">\ue000</span></span><span class="lsbc v1">y</span><span class="ff3 fs3 ls20 v1">)</span><span class="ls5a v1">t</span><span class="ff1 fs2 v9">2</span></div><div class="t m0 x7 h4 yb0 ff1 fs1 fc1 sc0 ls24 ws5f">Re-escreva esta exp<span class="_0 blank"></span>ressão como:</div><div class="t m0 x32 h4 yb1 ff1 fs1 fc1 sc0 ls24">1</div><div class="t m0 x34 h4a yb2 ff3 fs3 fc1 sc0 ls5">(<span class="ff1 fs1 lsf0 v0">2<span class="ff6 fs3 ls27">\ue000</span><span class="ff2 lsbc">y</span></span><span class="lsee">)<span class="ff2 fs1 ls24 ws5a v1">dy </span><span class="ls25 v1">=<span class="ff1 fs1 ls24 ws3">3<span class="ff2 lsd6">t</span><span class="fs2 ls36 v5">2</span><span class="ff2">dt</span></span></span></span></div><div class="t m0 x7 h4 yb3 ff1 fs1 fc1 sc0 ls24 ws21">Integrando amb<span class="_3 blank"> </span>os os lados, obtemos:</div><div class="t m0 x15 h4b yb4 ff5 fs7 fc1 sc0 lsf1">Z<span class="ff1 fs1 ls24 v20">1</span></div><div class="t m0 x26 h4c yb5 ff3 fs3 fc1 sc0 ls18">(<span class="ff1 fs1 lsf0 v0">2<span class="ff6 fs3 ls3">\ue000</span><span class="ff2 lsbc">y</span></span><span class="lsea">)<span class="ff2 fs1 ls24 ws5a v1">dy </span><span class="lsf2 v1">=</span><span class="ff5 fs7 ls2e v21">Z</span><span class="ff1 fs1 ls24 ws3 v1">3<span class="ff2 lse7">t<span class="ff1 fs2 ls36 v5">2</span><span class="ls24 ws60">dt <span class="ff4">,</span></span></span></span></span></div><div class="t m0 x7 h4 yb6 ff1 fs1 fc1 sc0 ls24 ws21">o que implica que:</div><div class="t m0 x36 h4d yb7 ff6 fs3 fc1 sc0 lsf3">\ue000<span class="ff1 fs1 ls24 ws61">ln </span><span class="lsf4">j<span class="ff1 fs1 lsf0">2</span><span class="ls3">\ue000<span class="ff2 fs1 lsbc">y</span><span class="lsf5">j<span class="ff3 ls80">=<span class="ff2 fs1 lsdc">t<span class="ff1 fs2 lsf6 v5">3</span></span><span class="ls7b">+<span class="ff2 fs1 ls24 ws3">k<span class="ff1 fs2 v7">1</span></span></span></span></span></span></span></div><div class="t m0 x28 h5 yb8 ff1 fs2 fc1 sc0 ls24 ws46">22</div></div><div class="c x0 y7 w2 h2"><div class="t m0 x6 h3 y8 ff1 fs0 fc0 sc0 ls24 ws4">Eq<span class="_0 blank"></span>ua<span class="_0 blank"></span>ções S<span class="_0 blank"></span>epa<span class="_b blank"></span>rá<span class="_0 blank"></span>vei<span class="_0 blank"></span>s de Pr<span class="_0 blank"></span>im<span class="_0 blank"></span>eira<span class="_0 blank"></span>-Or<span class="_0 blank"></span>dem</div><div class="t m0 x7 h4 yb9 ff1 fs1 fc1 sc0 ls24 ws3">(Cont.)</div><div class="t m0 x21 h34 yba ffa fsa fc0 sc0 ls81">I<span class="ff1 fs4 fc1 ls24 ws50 v7">Assim, temos que a solução do p<span class="_0 blank"></span>roblema acima é dada p<span class="_3 blank"> </span>or:</span></div><div class="t m0 x37 h4e ybb ff6 fs5 fc1 sc0 lsf7">j<span class="ff1 fs4 lsf8">2</span><span class="ls82">\ue000<span class="ff2 fs4 ls92">y</span><span class="lsf9">j<span class="ff3 lsb8">=<span class="ff2 fs4 lsfa">e</span></span><span class="fs8 lsfb v3">\ue000<span class="ff2 fs2 lsfc">t<span class="ff1 fsf lsfd v22">3</span></span><span class="lsfe v0">\ue000<span class="ff2 fs2 lsff">k<span class="ff1 fsf ls100 v23">1</span></span></span></span><span class="ls24 ws62 v0">$<span class="_1 blank"> </span>j<span class="ff1 fs4 lsf8">2</span><span class="ls82">\ue000<span class="ff2 fs4 ls92">y</span><span class="ls101">j<span class="ff3 lsd">=</span></span></span><span class="ff2 fs4 ws3c">ke </span><span class="fs8 lsfb v3">\ue000<span class="ff2 fs2 ls102">t<span class="ff1 fsf ls100 v22">3</span></span></span><span class="ls103">$<span class="ff1 fs4 ls104">2</span><span class="ls8a">\ue000<span class="ff2 fs4 ls6e">y</span><span class="ff3 ls6f">=</span><span class="ls105">\ue006</span></span></span><span class="ff2 fs4 ws3c">k<span class="_0 blank"></span>e <span class="ff6 fs8 lsfb v3">\ue000</span><span class="fs2 ls106 v3">t</span><span class="ff1 fsf v2">3</span></span></span></span></span></div><div class="t m0 x8 h4f ybc ff6 fs5 fc1 sc0 ls103">)<span class="ff2 fs4 ls107">y</span><span class="ff3 ls15 v0">(</span><span class="ff2 fs4 lsb">t</span><span class="ff3 lsc v0">)<span class="lsf v0">=<span class="ff1 fs4 lsf8">2</span><span class="ls108">+<span class="ff2 fs4 ls24 ws3c">k<span class="_0 blank"></span>e <span class="ff6 fs8 lsfb v3">\ue000</span><span class="fs2 ls109 v3">t</span><span class="ff1 fsf ls10a v2">3</span><span class="ff4">,</span></span></span></span></span></div><div class="t m0 x1f h1f ybd ff1 fs4 fc1 sc0 ls24 ws63">onde <span class="ff2 ls10b">k</span><span class="wsc">é uma constante a<span class="_0 blank"></span>rbitrária não-nula.</span></div><div class="t m0 x28 h5 y7d ff1 fs2 fc1 sc0 ls24 ws46">23</div></div></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div>
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