MHB Solve for $a^2$ in $u_{xx} = u_{tt}$ using $u_1$ and $u_2$

[karush]

1000
Find $a^2 u_{xx} = u_{tt} $ if $\lambda$ a real number.
If
$u_1\left(x, t\right)=\sin\left({\lambda x}\right)\sin\left({\lambda at}\right)$
And
$u_2 \left(x, t\right)={e}^{-a^2 \lambda ^2 t}\sin\left({\lambda x}\right)$

I didn't know how to deal with the $a^2$

 

[Prove It]

Find $a^2 u_{xx} = u_{tt} $ if $\lambda$ a real number.
If
$u_1\left(x, t\right)=\sin\left({\lambda x}\right)\sin\left({\lambda at}\right)$
And
$u_2 \left(x, t\right)={e}^{-a^2 \lambda ^2 t}\sin\left({\lambda x}\right)$

I didn't know how to deal with the $a^2$

What do you mean "find" this differential equation? Are you trying to solve it? Or are you trying to show that u1 and u2 are possible solutions?

 

[karush]

Yes sorry
Are $u^1$ and $u^2$ possible solutions
On the calculator I got
$${u}_{xx}=-\lambda^2\sin\left({\lambda x}\right) $$
So $\alpha$ disappeared

 

[Prove It]

Yes sorry
Are $u^1$ and $u^2$ possible solutions
On the calculator I got
$${u}_{xx}=-\lambda^2\sin\left({\lambda x}\right) $$
So $\alpha$ disappeared

It can't disappear. The entire $\displaystyle \begin{align*} \sin{ \left( \lambda \, a\, t \right) } \end{align*}$ term is a constant when differentiating partially with respect to x.

 

[karush]

Since $u_{xx}=-\lambda^2 \sin\left({\lambda x}\right)\sin\left({a\lambda t}\right)$

and

$u_{tt}=-a^2 \lambda^2 \sin\left({\lambda x}\right)\sin\left({a\lambda t}\right)$

Then $a^2 u_{xx}=u_{tt } $
I hope

 

[Prove It]

Since $u_{xx}=-\lambda^2 \sin\left({\lambda x}\right)\sin\left({a\lambda t}\right)$

and

$u_{tt}=-a^2 \lambda^2 \sin\left({\lambda x}\right)\sin\left({a\lambda t}\right)$

Then $a^2 u_{xx}=u_{tt } $
I hope

Yes that is correct :)

Can you do the other part of the question?