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I have a partial differential equation :

$u_{xx}-3u_{xt}-4u_{tt}=0$ with initial and boundary conditions: $u(x,0)=x^2$, $u_t(x,0)=e^x$. We need to find $u(x,t)$. The hint is : Factor the operator as was done with the wave equation $u_{tt}= c^2u_{xx}$.

Here is my thought process. Operator factoring is exactly like factoring from elementary school algebra. So I can pretend for a moment that

$u_{xx}-3u_{xt}-4u_{tt}=>(x^2-3xt-4t^2)=(x-4)(x+1) $

So

$u_{xx}-3u_{xt}-4u_{tt}= (\partial_x -4\partial_t)(\partial_x+\partial_t)u=0$

So the last step above is using the suggested hint of operator factoring.

From the general solution of wave equations we know it will have the form:

$u(x,t)=f(something1) + g(something2)$

Using the factored term: $(\partial_x -4\partial_t)$ we know that $something1 = x-4t$. Using the factored term $(\partial_x+\partial_t)$ we know that $something2 = x + t$.

Therefore the solution is:

$u(x,t)=f(x-4t) + g(x+t)$

Then I use the initial and boundary conditions giving me:

$u(x,0)=f(x)+g(x)=x^2$ $u_t(x,0)=g'(x)-4f'(x)=e^x$

However, I am unable to get the correct answer from here:

$\frac{4}{5} (e^{x+t/4}-e^{x-t}) + x^2+\frac{1}{4}t^2$

user1068636
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  • Possible duplicate: https://math.stackexchange.com/questions/1766614/solving-4u-tt-3u-xt-u-xx-0 https://math.stackexchange.com/questions/1406780/solve-the-pde-u-xx-3u-xt-4u-tt-0 https://math.stackexchange.com/questions/306250/solve-u-xx-3u-xt-4u-tt-0-where-ux-0-x2-and-u-tx-0-ex – K B Dave Mar 02 '18 at 20:53
  • From $u(x,0)=f(x)+g(x)=x^2$, $u_t(x,0)=g'(x)-4f'(x)=e^x$ you get $$g=x^{2}-f$$ So that $$f'=\frac{2}{5}x-\frac{1}{5}e^{x}$$ – Kiryl Pesotski Mar 02 '18 at 22:35
  • @user1068636 Correct answer is $u=\frac{4}{5} (e^{x+t/4}-e^{x-t}) + x^2+\frac{1}{4}t^2$. This can get and check with any CAS (Maxima, Maple, Mathematica, ...). – Aleksas Domarkas Jun 22 '18 at 08:53

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