Question

URGENTE: Determine a solução das equações de primeira ordem abaixo:

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Answer 1

$\large\boxed{\begin{array}{l}\rm a)~\sf y'+2ty=t^3\\\sf\dfrac{dy}{dt}+2ty=t^3\\~~~~\underline{\rm c\acute alculo\,do\,fator\,integrante\!:}\\\sf\mu(t)=e^{\displaystyle\int 2t\,dt}=e^{t^2}\\\\\sf \mu(t)\cdot\dfrac{dy}{dt}+\mu(t)\cdot 2ty=\mu(t)\cdot t^{3}\\\\\sf e^{t^2}\cdot\dfrac{dy}{dt}+e^{t^2}\cdot 2ty=e^{t^2}\cdot t^3\\\\\sf\dfrac{d}{dt}(e^{t^2}y)=t^3e^{t^2}\\\\\sf d(e^{t^2}y)=t^3e^{t^2}dt\\\displaystyle\sf\int d(e^{t^2}y)=\int t^3e^{t^2}\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf\int t^3~e^{t^2}dt=\dfrac{1}{2}\int t^2e^{t^2}2tdt\\\underline{\rm usando\,integral\,por\,partes\,temos:}\end{array}}$

$\large\boxed{\begin{array}{l}\sf u=t^2\implies du=2tdt\\\sf dv=e^{t^2}2tdt\implies v=e^{t^2}\\\displaystyle\sf\dfrac{1}{2}\int t^2e^{t^2}2tdt=\dfrac{1}{2}\bigg[ t^2e^{t^2}-\int e^{t^2}2tdt\bigg]\\\\\displaystyle\sf\dfrac{1}{2}\int t^2{e^{t^2}}2tdt=\dfrac{1}{2}\bigg[t^2e^{t^2}-e^{t^2}\bigg]+c_1\\\displaystyle\sf\dfrac{1}{2}\int t^2e^{t^2}2tdt=\dfrac{1}{2}e^{t^2}\bigg[t^2-1\bigg]+c_1\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm voltando\,a\,EDO\,original\,temos:}\\\displaystyle\sf\int d(e^{t^2}y)=\int t^3e^{t^2}dt\\\sf e^{t^2}y=\dfrac{1}{2}e^{t^2}(t^2-1)+c\\\\\sf y(t)=\dfrac{1}{2}\bigg(\dfrac{t^2-1}{e^{t^2}}\bigg)+\dfrac{c}{e^{t^2}}\end{array}}$