Resolução de Exercícios EDO

Prévia do material em texto

1)Considere a EDO: 
a) Mostre que essa EDO não é exata. 
 
M =x² +2x +y➔ My = 1 
N = 2 ➔ Nx = 0 
Não é exata 
 
b) Determine o fator integrante. Multiplique a EDO pelo fator 
integrante e mostre que a EDO obtida é exata. 
 
𝑀𝑦−𝑁𝑥
𝑁
 = 
1−0
2
=
1
2
=h(x) 
µ = 𝑒∫
1
2
𝑑𝑥 = 𝑒
1
2
𝑥 fator integrante 
 
(x² 𝑒
1
2
𝑥+2x𝑒
1
2
𝑥+y𝑒
1
2
𝑥 )𝑑𝑥 +2𝑒
1
2
𝑥𝑑𝑦 = 0 
M= x² 𝑒
1
2
𝑥+2x𝑒
1
2
𝑥+y𝑒
1
2
𝑥 ➔ My = 𝑒
1
2
𝑥 
N=2𝑒
1
2
𝑥 ➔ Nx =2. ½ 𝑒
1
2
𝑥 = 𝑒
1
2
𝑥 ➔ é exata 
 
c) Determine a solução geral dessa equação. 
𝑣 = ∫ 𝑒
1
2
𝑥𝑑𝑥 =
𝑒
1
2
𝑥
1
2
 =2𝑒
1
2
𝑥 
u=x² ➔ du = 2xdx 
∫ 𝑢𝑑𝑣 = 𝑢. 𝑣 − ∫ 𝑣𝑑𝑢 
 
F(x,y) = ∫(x2. 𝑒
1
2
𝑥 + 2x. 𝑒
1
2
𝑥 + y𝑒
1
2
𝑥 )𝑑𝑥 +g(y) 
F(x,y) = (x².2𝑒
1
2
𝑥 - ∫ 2𝑒
1
2
𝑥2𝑥𝑑𝑥) +(2x. 2𝑒
1
2
𝑥 -∫ 2𝑒
1
2
𝑥2𝑑𝑥) 
+y.2𝑒
1
2
𝑥+g(y) 
F(x,y) = (2x².𝑒
1
2
𝑥 - ∫ 4𝑥. 𝑒
1
2
𝑥𝑑𝑥) +(4x. 𝑒
1
2
𝑥 -∫ 4𝑒
1
2
𝑥𝑑𝑥) 
+2y.𝑒
1
2
𝑥+g(y) 
F(x,y) = 2x²𝑒
1
2
𝑥 – (4x. 2𝑒
1
2
𝑥 -∫ 2𝑒
1
2
𝑥4dx) +(4x. 𝑒
1
2
𝑥 -
4.2. 𝑒
1
2
𝑥 +2y.𝑒
1
2
𝑥+g(y) 
F(x,y) = 2x²𝑒
1
2
𝑥 – 8x. 𝑒
1
2
𝑥 +8. 2𝑒
1
2
𝑥+ 4x. 𝑒
1
2
𝑥 -8𝑒
1
2
𝑥 
+2y.𝑒
1
2
𝑥+g(y) 
F(x,y) = 2x²𝑒
1
2
𝑥 – 4x. 𝑒
1
2
𝑥 +8𝑒
1
2
𝑥+2y.𝑒
1
2
𝑥+g(y) 
 
Fy = N 
2𝑒
1
2
𝑥 +g’(y) = 2𝑒
1
2
𝑥 
g’(y) = 0 ➔ g(y) = ∫ 0𝑑𝑦 = 𝑐 
F(x,y) = 2x²𝑒
1
2
𝑥 – 4x. 𝑒
1
2
𝑥 +8𝑒
1
2
𝑥+2y.𝑒
1
2
𝑥 = 𝐶 
 
 d) Determine a solução particular para y(1) = 4. Y(x)=y 
2.1²𝑒
1
2
1 – 4.1. 𝑒
1
2
.1 +8𝑒
1
2
.1+2.4.𝑒
1
2
.1+ c = 0 ➔ c =- 14 √𝑒 
F(x,y) = 2x²𝑒
1
2
𝑥 – 4x. 𝑒
1
2
𝑥 +8𝑒
1
2
𝑥+2y.𝑒
1
2
𝑥= - 14 √𝑒 
 
2) Considere as EDOs lineares: 
a) 
3x
x
y
dx
dy
=− 
𝑑𝑦
𝑑𝑥
− 𝑥−1. 𝑦 = x³ ➔p(x) =-x-1 a.log b = log ba 
 
µ =𝑒∫ −𝑥
−1𝑑𝑥 = 𝑒−1.𝑙𝑛𝑥 = 𝑒𝑙𝑛𝑥
−1
= 𝑥−1 fator integrante 
 
𝑥−1
𝑑𝑦
𝑑𝑥
− 𝑥−1. 𝑥−1. 𝑦 =𝑥−1. x³ 
 
𝑥−1
𝑑𝑦
𝑑𝑥
− 𝑥−2. 𝑦 =𝑥 2 
Derivada do produto de µ com y 
 
 
𝑑
𝑑𝑥
(𝑥−1. 𝑦) = 𝑥² 
 
𝑥−1. 𝑦 = ∫ 𝑥² 𝑑𝑥 
𝑥−1. 𝑦 = 
𝑥3
3
+ 𝑐 ➔ 𝑦 =
𝑥3
3𝑥−1
+
𝑐
𝑥−1
 
 𝑦 =
𝑥4
3
 + c.x solução geral 
 
b) x²
dx
dy +(x²+4x)y = ex (:x²) 
𝑥2
𝑥2
𝑑𝑦
𝑑𝑥
+ (
𝑥2
𝑥2
+
4𝑥
𝑥2
) 𝑦 =
𝑒𝑥
𝑥²
 
𝑑𝑦
𝑑𝑥
+ (1 + 4𝑥−1)𝑦 = 𝑥−2𝑒𝑥 
 
𝜇 = 𝑒∫ 1+4𝑥
−1𝑑𝑥 = 𝑒𝑥+4𝑙𝑛𝑥 =ex . 𝑒𝑙𝑛𝑥
4
=ex .x4 ➔ fator 
integrante 
 
 
ex .x4 
𝑑𝑦
𝑑𝑥
+ (1 + 4𝑥−1)e𝑥 . x4𝑦 = 𝑥−2𝑒𝑥 ex .x4 
 
ex .x4 
𝑑𝑦
𝑑𝑥
+ (1 + 4𝑥−1)e𝑥 . x4𝑦 = 𝑥2𝑒2𝑥 
𝑑
𝑑𝑥
(𝑒𝑥𝑥4. 𝑦) = 𝑥2𝑒2𝑥 
 
𝑒𝑥𝑥4. 𝑦 = ∫ 𝑥2. 𝑒2𝑥 dx ➔**** 
 
 
∫ 𝑢𝑑𝑣 = 𝑢. 𝑣 − ∫ 𝑣𝑑𝑢 
 
∫ 𝑥2 𝑒2𝑥 dx = x².
𝑒2𝑥
2
− ∫
𝑒2𝑥
2
.2xdx 
 =
𝑥²𝑒2𝑥
2
− ∫ 𝑥. 𝑒2𝑥dx 
 =
𝑥²𝑒2𝑥
2
 – (x. 
𝑒2𝑥
2
− ∫
𝑒2𝑥
2
𝑑𝑥) 
=
𝑥²𝑒2𝑥
2
 – 
𝑥𝑒2𝑥
2
+ 
𝑒2𝑥
4
 + c 
 
**** voltando ao exercício 
𝑒𝑥𝑥4. 𝑦 =
𝑥²𝑒2𝑥
2
 – 
𝑥𝑒2𝑥
2
+ 
𝑒2𝑥
4
 + c 
𝑦 = 
𝑥²𝑒2𝑥
2𝑒𝑥𝑥4.
 – 
𝑥𝑒2𝑥
2𝑒𝑥𝑥4.
+ 
𝑒2𝑥
4𝑒𝑥𝑥4.
 + 
𝑐
𝑒𝑥𝑥4
 
 
𝑦 = 
𝑒𝑥
2𝑥2
 – 
𝑒𝑥
2𝑥3
+ 
𝑒𝑥
4𝑥4.
 + 
𝑐
𝑒𝑥𝑥4
 solução geral 
 
 
c) 
²
1
²
5
'
x
y
x
y =− 
𝑑𝑦
𝑑𝑥
− 5𝑥−2. 𝑦 = x-2 ➔p(x) =-5x-2 
 
µ =𝑒∫ −5𝑥
−2𝑑𝑥 = 𝑒−
5𝑥−1
−1 = 𝑒5𝑥
−1
 fator integrante 
 
𝑒5𝑥
−1 𝑑𝑦
𝑑𝑥
− 𝑒5𝑥
−1
. 5𝑥−2. 𝑦 =𝑒5𝑥
−1
𝑥−2 
Derivada do produto de µ com y 
 
 
𝑑
𝑑𝑥
(𝑒5𝑥
−1
. 𝑦) = 𝑒5𝑥
−1
𝑥−2 
 
𝑒5𝑥
−1
. 𝑦 = ∫ 𝑒5𝑥
−1
𝑥−2 𝑑𝑥 
u =x² ➔ du = 2xdx 
dv= e2x dx ➔ v= 
𝑒2𝑥
2
 
u=x➔du =1dx 
dv= e2x dx ➔ v= 
𝑒2𝑥
2
 
𝑒5𝑥
−1
. 𝑦 = 
𝑒5𝑥
−1
.𝑥−2
−5𝑥−2
=
𝑒5𝑥
−1
−5
+ 𝑐 
𝑦 =
𝑒5𝑥
−1
−5𝑒5𝑥
−1 +
𝑐
𝑒5/𝑥
 
 
 𝑦 =
−1
5
 + 
𝑐
𝑒
5
𝑥
 solução geral 
 
 
3)Resolva por variáveis separáveis: 
A) y’ = x²y com y(1) = 3. 
𝑑𝑦
𝑑𝑥
= 𝑥2𝑦 
𝑑𝑦 = 𝑥2𝑦𝑑𝑥 
𝑑𝑦
𝑦
= 𝑥2𝑑𝑥 
𝑦−1 𝑑𝑦 = 𝑥2𝑑𝑥 
∫ 𝑦−1 𝑑𝑦 = ∫ 𝑥2 𝑑𝑥 
𝑙𝑛𝑦 =
𝑥3
3
+ 𝑐 
𝑒𝑙𝑛𝑦 =𝑒
𝑥³
3
+𝑐 
𝑦 = 𝑒
𝑥³
3 . 𝑒𝑐➔ 𝑦 = 𝑒
𝑥³
3 . 𝑘 ==> 𝑦 = 𝑘. 𝑒
𝑥³
3 solução geral 
Solução particular 3 = 𝑘. 𝑒
13
3 ➔ 𝑘 =
3
𝑒
1
3
 =
3
√𝑒
3 
𝑦 =
3
𝑒
1
3
 . 𝑒
𝑥³
3 
y = 3. 𝑒
𝑥3−1
3 
B) 
𝑑𝑦
𝑑𝑥
=
2𝑦−1
𝑥+1
 
𝑑𝑦
2𝑦 − 1
=
𝑑𝑥
𝑥 + 1
 
(2𝑦 − 1)−1𝑑𝑦 = (𝑥 + 1)−1 𝑑𝑥 
 
∫(2𝑦 − 1)−1𝑑𝑦 = ∫(𝑥 + 1)−1 𝑑𝑥 
ln(2𝑦−1)
2
 = ln(𝑥 + 1) + 𝑐 
ln(2𝑦 − 1) = 2 ln(𝑥 + 1) + 2𝑐 
𝑒ln (2𝑦−1) = 𝑒ln(𝑥+1)
2+2𝑐 
2𝑦 − 1 = 𝑒ln(𝑥+1)
2
𝑒2𝑐 
2𝑦 − 1 = (𝑥 + 1)2. 𝑘 
 𝑦 =
𝑘(𝑥+1)2+1
2
 solução geral 
 
c) y
dx
dy
x 6= ➔ 
xdy = 6y dx 
𝑑𝑦
6𝑦
=
𝑑𝑥
𝑥
 
𝑦−1𝑑𝑦 = 6𝑥−1𝑑𝑥 
∫ 𝑦−1𝑑𝑦 = ∫ 6𝑥−1𝑑𝑥 
𝑙𝑛𝑦 = 6𝑙𝑛𝑥 + 𝑐 
𝑒𝑙𝑛𝑦 = 𝑒𝑙𝑛𝑥
6+𝑐 
𝑦 = 𝑒𝑙𝑛𝑥
6
. 𝑒𝑐 
𝑦 = 𝑥6.k solução geral 
d) 
21
2
x
xy
dx
dy
−
= 
(1 − 𝑥2)𝑑𝑦 = 2𝑥𝑦𝑑𝑥 
𝑑𝑦
𝑦
=
2𝑥
1 − 𝑥2
 𝑑𝑥 
∫ 𝑦−1 𝑑𝑦 = ∫ 2𝑥. (1 − 𝑥2)−1 𝑑𝑥 
𝑙𝑛 𝑦 =
2𝑥𝑙𝑛(1−𝑥2)
−2𝑥
 + c 
lny = - ln(1-x²)+ c 
𝑒𝑙𝑛𝑦 = 𝑒− ln(1−𝑥
2)+𝑐 
𝑦 = 𝑒ln(1−𝑥
2)
−1
. 𝑒𝑐 
𝑦 = (1 − 𝑥2)−1. 𝑘 
𝑦 =
𝑘
1−𝑥2
 solução geral 
e) 4)3(;
5
=
−
= y
x
y
dx
dy 
𝑑y
y
=
dx
5 − x
 
𝑦−1dy = (5 − x)−1dx 
∫ 𝑦−1 dy = ∫(5 − x)−1dx 
𝑙𝑛𝑦 =
ln(5 − 𝑥)
−1
+ 𝑐 
𝑒𝑙𝑛𝑦 = 𝑒ln (5−𝑥)
−1+𝑐 
𝑦 = (5 − 𝑥)−1. 𝑘 
𝑦 =
𝑘
5−𝑥
 Solução geral 
4 =
𝑘
5−3
➔ k=8 
𝑦 =
8
5−𝑥
 solução particular 
 
 
5) Verifique se a Edo é homogênea (diga o grau) e resolva: 
a) 
x
yx
y
23
'
+
= 
 
𝑑𝑦
𝑑𝑥
=
3𝑥+2𝑦
𝑥
 
𝑥𝑑𝑦 = (3𝑥 + 2𝑦)𝑑𝑥 
(3𝑥 + 2𝑦)𝑑𝑥 − 𝑥𝑑𝑦 = 0 
M = 3x+2y → grau 1 
N = -x → grau 1 
Homogênea de grau 1 
y= t.x ➔ dy = tdx+xdt➔ t =
𝑦
𝑥
 (usa no final) 
 
(3𝑥 + 2𝑦)𝑑𝑥 − 𝑥𝑑𝑦 = 0 
 
(3𝑥 + 2𝑡𝑥)𝑑𝑥 − 𝑥(tdx + xdt) = 0 
3𝑥𝑑𝑥 + 2𝑡𝑥𝑑𝑥 − 𝑡𝑥𝑑𝑥 − 𝑥2𝑑𝑡 = 0 
3𝑥𝑑𝑥 + 𝑡𝑥𝑑𝑥 − 𝑥2𝑑𝑡 = 0 
(3 + 𝑡)𝑥𝑑𝑥 = 𝑥2𝑑𝑡 
𝑥𝑑𝑥
𝑥2
=
𝑑𝑡
3 + 𝑡
 
𝑥−1𝑑𝑥 = (3 + 𝑡)−1𝑑𝑡 
∫ 𝑥−1𝑑𝑥 = ∫(3 + 𝑡)−1𝑑𝑡 
 
𝑙𝑛𝑥 = ln(3 + 𝑡) + 𝑐 
𝑒𝑙𝑛𝑥 = 𝑒ln(3+𝑡)+𝑐 
𝑥 = 𝑒ln(3+𝑡). 𝑒𝑐 
𝑥 = 𝑘. (3 + 𝑡) 
𝑥 = 𝑘. (3 +
𝑦
𝑥
) 
𝑥 = 𝑘
(3𝑥 + 𝑦)
𝑥
 
𝑥2 = 3𝑥𝑘 + 𝑘𝑦 
𝑦 =
𝑥2
𝑘
− 3𝑥 
𝑦 = 𝑐𝑥² − 3𝑥 solução geral 
b) (x + y) dx+ (x-y) dy = 0 
M= x + y ➔ grau 1 
N = x - y ➔ grau 1 EDO homogênea grau 1 
 
 y = t.x ➔ dy = t.dx+x.dt ➔ t = 
𝑦
𝑥
 ( usa no final) 
 
(x + y) dx+ (x-y) dy = 0 
(x+tx)dx+(x-tx)(tdx+xdt)=0 
 
xdx+txdx+txdx+x²dt-t²xdx-tx²dt=0 
xdx+txdx+txdx+x²dt-t²xdx-tx²dt=0 
 (1+2t-t²) xdx + (1-t) x²dt=0 
 
 (1+2t-t²) x dx = - (1-t) x²dt 
−
𝑥. 𝑑𝑥
𝑥2
 =
1 − 𝑡
1 + 2𝑡 − 𝑡2
𝑑𝑡 
 
∫ −𝑥−1𝑑𝑥 = ∫(1 − 𝑡)(1 + 2𝑡 − 𝑡2)−1𝑑𝑡 
-lnx = 
(1−𝑡)𝐿𝑛(1+2𝑡−𝑡2)
2−2𝑡
+ 𝑐 
 
-lnx = 
(1−𝑡)𝐿𝑛(1+2𝑡−𝑡2)
2(1−𝑡)
+ 𝑐 
- lnx = 
𝑙𝑛(1+2𝑡−𝑡2)
2
+ 𝑐 
-2 . lnx =ln(1+2t-t²)+c 
ln x-2 = ln(1+2t-t²)+c 
𝑒ln x
−2 =𝑒ln(1+2𝑡−𝑡
2)+𝑐 
𝑒ln x
−2 =𝑒ln(1+2𝑡−𝑡
2).𝑒c 
x-2 =(1+2t-t²).k 
1
𝑥2
 =(1+ 
2𝑦
𝑥
 – 
𝑦²
𝑥²
).k 
1
𝑘𝑥2
=
𝑥2+2𝑥𝑦−𝑦²
𝑥²
 
1
𝑘
= 𝑥2 + 2𝑥𝑦 − 𝑦2 
𝐶 = 𝑥2 + 2𝑥𝑦 − 𝑦2 
 
 
6) Verifique se as Edo’s são exatas e resolva: 
a) (4x³y-ln x)dx +(x4- 2ln y)dy = 0 
 
M=4x³y-ln x➔ My=4x³ 
N= x4- 2ln y ➔ Nx= 4x³ 
Edo é exata 
 
F(x,y) =∫ 4x3y − ln xdx + 𝑔(𝑦) 
F(x,y) = 
4x4
4
y − x. lnx + x + 𝑔(𝑦) 
F(x,y) = x4y − x. lnx + x + 𝑔(𝑦) 
Fy = N 
x4 +g’(y) = x4- 2ln y 
𝑔′(𝑦) = −2𝑙𝑛𝑦 
g(y) =∫ −2𝑙𝑛𝑦 𝑑𝑦 = −2(𝑦𝑙𝑛𝑦 − y) + 𝑐 
 
F(x,y) = x4y − x. lnx + x − 2𝑦𝑙𝑛𝑦 + 2𝑦 + 𝑐 = 0 
 
b) (y+2xy³)dx+(1+3x²y²+x)dy=0 
 
M=y+2xy³ ➔ My = 1+6xy² 
N=1+3x²y²+x➔ Nx = 6xy²+1 
Edo exata 
 
F(x,y) =∫ y + 2xy3dx = yx +
2x2
2
y³ + 𝑔(𝑦) 
F(x,y) = yx + 𝑥²y³ + 𝑔(𝑦) 
Fy = N 
1x +3x²y²+g’(y) = 1+3x²y²+x 
𝑔′(𝑦) = 1 
g(y) =∫ 1𝑑𝑦 = 𝑦 + 𝑐 
 
F(x,y) = yx + 𝑥2y3 + 𝑦 + 𝑐 = 0 
F(x,y) = yx + 𝑥2y3 + 𝑦 = 𝐶