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I looked for similar questions and I found this one and this one;
But these two questions are about another method. I want to solve the following problem with Separation of Variables.
$$ \begin{cases} u_{tt}=u_{xx}\\ u(x,0)=x \\ u_t(x,0)=0 \\ u_x(0,t)=2\sin(t)\\ u_x(\pi,t)=0 \end{cases} $$

I know how to solve the Homogeneous Wave Equations with Homogeneous B.Cs, but I have no idea for this problem.

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I know I have to write $u(x,t)=X(x)T(t)$, but I don't know how to complete the answer: $$ u(x,t)=X(x)T(t) \quad\Rightarrow \quad XT''=X''T $$ $$ \Rightarrow\quad \frac{X''}{X}=\frac{T''}{T}=\lambda \quad \Rightarrow$$ \begin{cases} X''-\lambda X=0 \\ X'(0)T(t)=2\sin(t)\\ X'(\pi)=0 \end{cases} Now from the $X'(0)T(t)=2\sin(t)$ what we get? $$ X'(0)T(t)=2\sin(t) \Longrightarrow T(t)\int X'(x)dx|_{x=0} = \int 2\sin(t)dx$$ $$ \Rightarrow X(x)|_{x=0}T(t)=2x\sin(t) $$ It looks not to be correct! I mean, is $X(0)=0$ ?

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1 Answers1

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Hint: If \begin{cases} u_x(0,t)=p\\ u_x(\pi,t)=q \end{cases} we at first make vanish $p$ and $q$ with another assumption. Let $u(x,t)=v(x,t)+w(x,t)$ so for $w(x,t)=\alpha x+\beta$, from \begin{cases} u_x(0,t)=2\sin t\\ u_x(\pi,t)=0 \end{cases} we have $w(x,t)=(-\dfrac{2}{\pi}\sin t)x+2\sin t$ and from $u(x,0)=v(x,0)+w(x,0)$ we know $v(x,0)=x$, also with $u_t(x,0)=v_t(x,0)+w_t(x,0)$ then $v_t(x,0)=\dfrac{2}{\pi}x-2$, so new problem takes the form \begin{cases} v_{tt}=v_{xx}+\left(-\dfrac{2}{\pi}x+2\right)\sin t\\ v(x,0)=x \\ v_t(x,0)=\dfrac{2}{\pi}x-2 \\ v_x(0,t)=0\\ v_x(\pi,t)=0 \end{cases}

Nosrati
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