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[SOLVED] a^2 U_xx = u_tt

karush

Well-known member
Jan 31, 2012
2,928
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 $a^2$
 
Last edited:

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,430
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

Well-known member
Jan 31, 2012
2,928
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
 
Last edited:

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,430
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

Well-known member
Jan 31, 2012
2,928
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

Well-known member
MHB Math Helper
Jan 26, 2012
1,430
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?