Question

Verifique se as equações diferencial a seguir é ordinária ou parcial e se a função dada é solução da equação.

Answer 1

✅ Ao identificar as equações e testar suas possíveis soluções, obtemos:

 

✍ Solução:

$\large\begin{array}{lr}\rm a)\\\rm {\underbrace{\rm u_{xx}+u_{yy}}_{\star}=0\,,~u(x,y)=\ln(x^2 + y^2)}\\\\\rm i)~\dfrac{\partial u}{\partial x}=\dfrac{2x}{x^2+y^2}\Rightarrow2\cdot\dfrac{\partial^2u}{\partial x^2}=\dfrac{x^2+y^2-x(2x) }{(x^2+y^2)^2}=2\cdot\dfrac{x^2+y^2-2x^2}{(x^2+y^2)^2}=-\dfrac{2(x^2-y^2)}{(x^2+y^2)^2}\\\\\rm ii)~\dfrac{\partial u}{\partial y}=\dfrac{2y}{x^2+y^2}\Rightarrow2\cdot\dfrac{\partial^2u}{\partial y^2}=\dfrac{x^2+y^2-y(2y)}{(x^2+y^2)^2}=2\cdot\dfrac{x^2+y^2-2y^2}{(x^2+y^2)^2}=\dfrac{2(x^2-y^2)}{(x^2+y^2)^2}\\\\\rm\underbrace{\raisebox{}{ \rm-\dfrac{2(x^2-y^2)}{(x^2+y^2)^2}+\dfrac{2(x^2-y^2)}{(x^2+y^2)^2} }}_{\star}=0\\\\\red{\underline{\boxed{\boxed{\rm \therefore\: u(x,y)~\acute{e}~sol.~da~EDP}}}}\end{array}$

 

$\large\begin{array}{lr}\rm b)\\\rm\ddot{y}-y=0\,,~y(t)=e^t\\\\\rm~e^t-e^t=0\\\\\red{\underline{\boxed{\boxed{\rm \therefore\:y(t)\acute{e}~sol.~da~EDO}}}}\end{array}$

 

$\large\begin{array}{lr}\rm c)\\\rm\ddot{y}-y=0\,,~y(t)=\cosh(t)\\\\\rm i:~\dfrac{dy}{dt}=\sinh(t)\Rightarrow\dfrac{d^2y}{dt^2}=\cosh(t)\\\\\rm ii:~\cosh(t)-\cosh(t)=0\\\\\rm\dfrac{e^x-e^{-x}}{2}-\dfrac{e^x-e^{-x}}{2}=0\\\\\red{\underline{\boxed{\boxed{\rm \therefore\:y(t)~\acute{e}~sol.~da~EDO}}}} \end{array}$

 

$\large\begin{array}{lr}\rm d)\\\rm {\large\underbrace{\rm u_{xx}+u_{yy}}_{\star}=0\,,~u(x,y)=\cos(x)\cosh(y)}\\\\\rm i:~\dfrac{\partial u}{\partial x}=-\sin(x)\cosh(y)\Rightarrow\dfrac{\partial u^2}{\partial x^2}=-\cos(x)\cosh(y)\\\\\rm ii:~\dfrac{\partial u}{\partial y}=\cos(x)\sinh(y)\Rightarrow\dfrac{\partial u^2}{\partial y^2}=\cos(x)\cosh(y)\\\\\rm{\large\underbrace{\rm-\cos(x)\cosh(y)+\cos(x)\cosh(y)}_{\large\star}=0}\\\\\red{\underline{\boxed{\boxed{\rm \therefore\:u(x,y)~\acute{e}~sol.~da~EDP}}}} \end{array}$

 

$\large\begin{array}{lr}\rm e)\\\rm t\dot{y}-y=t^2\,,~y(t)=3t+t^2\\\\\rm i:~\dfrac{dy}{dt} = 3+2t\\\\\rm ii:~ t(3+2t)-(3t+t^2)=t^2\\\\\rm 3t+2t^2-3t-t^2=t^2\\\\\rm t^2 = t^2\\\\\red{\underline{\boxed{\boxed{\rm \therefore\:y(t)~\acute{e}~sol.~da~EDO}}}} \end{array}$

 

Questão d) no anexo