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Encontrar a solução em série da EDO abaixo ?

y"+y=0

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Escrevendo y e y'' em series de potencias, temos:


y=\sum_{n=0}^{+\infty}a_nx^n\\
y'=\left(\sum_{n=0}^{+\infty}a_nx^n\right)'=\sum_{n=0}^{+\infty}(a_nx^n)'=
\sum_{n=0}^{+\infty}a_nnx^{n-1}=\\\sum_{n=1}^{+\infty}a_nnx^{n-1}\equiv\sum_{n=0}^{+\infty}a_{n+1}(n+1)x^n\\
y''=(y')'=\left[\sum_{n=0}^{+\infty}a_{n+1}(n+1)x^n\right]'=\sum_{n=0}^{+\infty}\left[a_{n+1}(n+1)x^n\right]'=\\
\sum_{n=0}^{+\infty}a_{n+1}(n+1)nx^{n-1}=\sum_{n=1}^{+\infty}a_{n+1}(n+1)nx^{n-1}\equiv\sum_{n=0}^{+\infty}a_{n+2}(n+2)(n+1)x^n


Colocando na igualdade:


(1-x)y''=y\\
(1-x)\sum_{n=0}^{+\infty}a_{n+2}(n+2)(n+1)x^n=\sum_{n=0}^{+\infty}a_nx^n\\
\sum_{n=0}^{+\infty}a_{n+2}(n+2)(n+1)x^n-x\sum_{n=0}^{+\infty}a_{n+2}(n+2)(n+1)x^n=\sum_{n=0}^{+\infty}a_nx^n\\
\sum_{n=0}^{+\infty}a_{n+2}(n+2)(n+1)x^n-\sum_{n=0}^{+\infty}a_{n+2}(n+2)(n+1)x^{n+1}=\sum_{n=0}^{+\infty}a_nx^n


Comparando x com xx^2 com x^2 e manipulando a expressão transformando o segundo somatório, temos:


\sum_{n=0}^{+\infty}a_{n+2}(n+2)(n+1)x^n-\sum_{n=0}^{+\infty}a_{n+2}(n+2)(n+1)x^{n+1}=\sum_{n=0}^{+\infty}a_nx^n\\
\sum_{n=0}^{+\infty}a_{n+2}(n+2)(n+1)x^n-\sum_{n=0}^{+\infty}a_{n+1}(n+1)nx^{n}=\sum_{n=0}^{+\infty}a_nx^n


Portanto,


a_{n+2}(n+2)(n+1)-a_{n+1}(n+1)n=a_n,n=0,1,2,3,4,\cdots\\
a_{n+2}=\dfrac{a_n+a_{n+1}(n+1)n}{(n+2)(n+1)}

Escrevendo y e y'' em series de potencias, temos:


y=\sum_{n=0}^{+\infty}a_nx^n\\
y'=\left(\sum_{n=0}^{+\infty}a_nx^n\right)'=\sum_{n=0}^{+\infty}(a_nx^n)'=
\sum_{n=0}^{+\infty}a_nnx^{n-1}=\\\sum_{n=1}^{+\infty}a_nnx^{n-1}\equiv\sum_{n=0}^{+\infty}a_{n+1}(n+1)x^n\\
y''=(y')'=\left[\sum_{n=0}^{+\infty}a_{n+1}(n+1)x^n\right]'=\sum_{n=0}^{+\infty}\left[a_{n+1}(n+1)x^n\right]'=\\
\sum_{n=0}^{+\infty}a_{n+1}(n+1)nx^{n-1}=\sum_{n=1}^{+\infty}a_{n+1}(n+1)nx^{n-1}\equiv\sum_{n=0}^{+\infty}a_{n+2}(n+2)(n+1)x^n


Colocando na igualdade:


(1-x)y''=y\\
(1-x)\sum_{n=0}^{+\infty}a_{n+2}(n+2)(n+1)x^n=\sum_{n=0}^{+\infty}a_nx^n\\
\sum_{n=0}^{+\infty}a_{n+2}(n+2)(n+1)x^n-x\sum_{n=0}^{+\infty}a_{n+2}(n+2)(n+1)x^n=\sum_{n=0}^{+\infty}a_nx^n\\
\sum_{n=0}^{+\infty}a_{n+2}(n+2)(n+1)x^n-\sum_{n=0}^{+\infty}a_{n+2}(n+2)(n+1)x^{n+1}=\sum_{n=0}^{+\infty}a_nx^n


Comparando x com xx^2 com x^2 e manipulando a expressão transformando o segundo somatório, temos:


\sum_{n=0}^{+\infty}a_{n+2}(n+2)(n+1)x^n-\sum_{n=0}^{+\infty}a_{n+2}(n+2)(n+1)x^{n+1}=\sum_{n=0}^{+\infty}a_nx^n\\
\sum_{n=0}^{+\infty}a_{n+2}(n+2)(n+1)x^n-\sum_{n=0}^{+\infty}a_{n+1}(n+1)nx^{n}=\sum_{n=0}^{+\infty}a_nx^n


Portanto,


a_{n+2}(n+2)(n+1)-a_{n+1}(n+1)n=a_n,n=0,1,2,3,4,\cdots\\
a_{n+2}=\dfrac{a_n+a_{n+1}(n+1)n}{(n+2)(n+1)}

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Hertz

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y''(x) = -y(x)

Solve ( d^2 y(x))\/( dx^2) = -y(x):\nAssume a solution will be proportional to e^(lambda x) for some constant lambda.\nSubstitute y(x)  =  e^(lambda x) into the differential equation:\n( d^2 )\/( dx^2)(e^(lambda x))+e^(lambda x)  =  0\nSubstitute ( d^2 )\/( dx^2)(e^(lambda x))  =  lambda^2 e^(lambda x):\nlambda^2 e^(lambda x)+e^(lambda x)  =  0\nFactor out e^(lambda x):\n(lambda^2+1) e^(lambda x)  =  0\nSince e^(lambda x) !=0 for any finite lambda, the zeros must come from the polynomial:\nlambda^2+1  =  0\nSolve for lambda:\nlambda = i or lambda = -i\nThe roots lambda  =  ± i give y_1(x) = c_1 e^(i x), y_2(x) = c_2 e^(-i x) as solutions, where c_1 and c_2 are arbitrary constants.\nThe general solution is the sum of the above solutions:\ny(x)  =  y_1(x)+y_2(x)  =  c_1 e^(i x)+c_2 e^(-i x)\nApply Euler\'s identity e^(alpha+i beta) = e^alpha cos(beta)+i e^alpha sin(beta):\ny(x)  =  c_1 (cos(x)+i sin(x))+c_2 (cos(x)-i sin(x))\nRegroup terms:\ny(x)  =  (c_1+c_2) cos(x)+i (c_1-c_2) sin(x)\nRedefine c_1+c_2 as c_1 and i (c_1-c_2) as c_2, since these are arbitrary constants:\nAnswer: |  \n | y(x)  =  c_1 cos(x)+c_2 sin(x)

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