Question

Resolver a EQ diferencial Linear​

Answer 1

$\boxed{\begin{array}{l}\underline{\rm equac_{\!\!,}\tilde ao~diferencial~linear}\\\sf D~\!\!efinic_{\!\!,}\tilde ao:~\dfrac{dy}{dx}+p(x)\cdot y=q(x)\\\displaystyle\sf Fator~integrante:~\mu(x)=e^{\int p(x)~dx}\end{array}}$

$\boxed{\begin{array}{l}\sf y'=2y+x^2+5\\\sf\dfrac{dy}{dx}=2y+x^2+5\\\sf\dfrac{dy}{dx}-2y=x^2+5\\\underline{\rm c\acute alculo~do~fator~integrante:}\\\displaystyle\sf\mu(x)=e^{\int p(x)~dx}\\\displaystyle\sf \mu(x)=e^{\int -2~dx}=e^{-2x}\\\underline{\rm multiplicando~a~equac_{\!\!,}\tilde ao~diferencial}\\\underline{\rm pelo~fator~integrante~temos:}\end{array}}$

$\boxed{\begin{array}{l}\sf e^{-2x}\cdot\dfrac{dy}{dx}-e^{-2x}\cdot 2y=e^{-2x}(x^2-5)\\\sf\dfrac{d}{dx}(e^{-2x}\cdot y)=x^2e^{-2x}+5e^{-2x}\\\sf d(e^{-2x}\cdot y)=(x^2e^{-2x}+5e^{-2x})d x\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm integrando~dos~dois~lados~temos:}\\\displaystyle\sf\int d(e^{-2x}y)=\int(x^2e^{-2x}+5e^{-2x})dx\\\sf ye^{-2x}=-\dfrac{1}{2}x^2e^{-2x}-\dfrac{1}{2}xe^{-2x}-\dfrac{1}{4}e^{-2x}-\dfrac{5}{2}e^{-2x}+k\\\sf y=-\dfrac{1}{2}x^2-\dfrac{1}{2}x-\dfrac{1}{4}-\dfrac{5}{2}+ke^{2x}\\\sf y(x)=-\dfrac{1}{2}x^2-\dfrac{1}{2}x-\dfrac{11}{4} ke^{2x}\end{array}}$