# Thread: 1.6 Verify the differential equation is = to the solution

1. ## 1.6 Verify the differential equation is = to the solution

Verify for each of the following differential equations that given function or functions are solutions:

$\textbf{(a)}$
\begin{align*}\displaystyle
u_{xx}+u_{yy}&=0\\
u_1(x,y)&=x^2-y^2\\
u_2(x,y)&=\cos{x}cosh{y}
\end{align*}
$\textbf{(b)$\lambda$a real constant }$
\begin{align*}\displaystyle
a^2u_{xx}&=u_t\\
u_1(x,t)&=e^{-\alpha^2t}\sin{x}\\
u_2(x,t)&=e^{-\alpha^2\lambda^2t}\sin{\lambda x}
\end{align*}
$\textbf{(c)$\lambda$a real constant }$
\begin{align*}\displaystyle
a^2u_{xx}&=u_{tt}\\
u_1(x,t)&=\sin{\lambda x}\sin{\lambda \alpha t}\\
u_2(x,t)&=\sin{(x-\alpha t)}
\end{align*}

ok I have never done this before and just trying to learn as much as I can before I take the class
I read the chapter but it was kinda ???

2. ## Re: 1.6 Verify the differential equation is = to the solution

Originally Posted by bigwave
Verify for each of the following differential equations that given function or functions are solutions:

$\textbf{(a)}$
\begin{align*}\displaystyle
u_{xx}+u_{yy}&=0\\
u_1(x,y)&=x^2-y^2\\
u_2(x,y)&=\cos{x}cosh{y}
\end{align*}

${u_1}_x =2x;~{u_1}_{xx} = 2$

${u_1}_y = -2y;~{u_1}_{yy} = -2$

${u_1}_{xx} + {u_1}_{yy} = 2+(-2) = 0$

similarly

${u_2}_{xx} = -\cos(x)\cosh(y)$

${u_2}_{yy} = \cos(x)\cosh(y)$

${u_2}_{xx} + {u_2}_{yy} = 0$

you can figure out the rest

3. ## Re: 1.6 Verify the differential equation is = to the solution

so what is $a^2$ and $u_t$

in (b) and (c)

4. ## Re: 1.6 Verify the differential equation is = to the solution

Originally Posted by bigwave
so what is $a^2$ and $u_t$

in (b) and (c)
$a^2$ is just some constant

oh I see the confusion. I bet they meant to use either just $a$ or just $\alpha$. Work the problem assuming $a=\alpha$ and see if it works out.

$u_t = \dfrac{du}{dt}$

5. ## Re: 1.6 Verify the differential equation is = to the solution

ok i think alpha in all cases