given: $\alpha, \beta$ are constants. find a solution to the following pde in the form $u(x,t) = B(t)\cdot e^{\alpha x + \beta t}$ dependant on their values: $$u_{tt} -c^2 u_{xx} = e^{\alpha x + \beta t}$$ the solution given is: $$\begin{cases} B(t) = \frac{1}{\beta ^2 - c^2\alpha ^2} \quad \alpha ^2 \neq \beta ^2 \\ B(t) = \frac{t}{2\beta} \quad \text{else} \end{cases}$$ I attempted to solve algebriacally by subbing in the attempted solution to the formula then got stuck on this expression:$$B(t)\beta + B'(t)(2\beta - \alpha) + B''(t) = 1$$ and am unsure at all as to how to continue. tried using $B(t) = e^{rt}$ but that led me nowhere, I was hoping you could please help me out as I've been stuck on this for a while. thanks.
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1$\begingroup$ \begin{align} u_{t} &= (B' + \beta B) e^{\alpha x + \beta t} \\ u_{tt} &= (B'' + 2 \beta B' + \beta^{2} B) e^{\alpha x + \beta t} \\ u_{xx} &= \alpha^{2} B e^{\alpha x + \beta t} \\ \implies u_{tt} - c^{2} u_{xx} &= \color{red}{(B'' + 2 \beta B' + \beta^{2} B - c^{2} \alpha^{2} B)} e^{\alpha x + \beta t} \\ &= e^{\alpha x + \beta t} \end{align} $\endgroup$– Matthew CassellMay 15, 2022 at 3:40
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1 Answer
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It looks like your resulting ODE is incorrect. You should have a term with $\beta^2$ and another term with $\alpha^2$ since you're differentiating $u$ twice with respect to each variable.
To actually solve nonhomogeneous linear 2nd order ODEs there's a number of different methods, and usually you'll have to first find the homogeneous solution, i.e. solve the ODE with right hand side $=0$. This essentially consists of making the ansatz $B(t)=e^{rt}$ and finding the roots of the resulting quadratic polynomial. Then use a method like variation of parameters or method of undetermined coefficients to find the particular solution, which is added to the homogeneous solution to produce the general solution.
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1$\begingroup$ I'll redo it and continue as suggested, I'll make an edit if it won't work. thank you! $\endgroup$ May 14, 2022 at 17:38