Great! Since there may be others who are wondering, here is how I would attempt this problem:
As we do in ordinary differential equations with a "linear non-homogeneous equation", look for the general solution to the associated "homogeneous equation, \(\displaystyle u_t= u_{xx}- u\). Try separating variables- that is, look for a solution of the form \(\displaystyle u(x, t)= X(x)T(t)\) where X is a function of X only and T is a function of t only. The differential equation becomes \(\displaystyle XT'= X''T- XT- xe^{-t}\). If divide by u= XT, that becomes \(\displaystyle \frac{T'}{T}= \frac{X''}{X}+ 1\). The left side of the equation depends on "x" only while the right side depends on "t" only. Since they must be equal for all x and t, each side must be equal to a constant.
We can separate the equation into \(\displaystyle \frac{T'}{T}= \alpha\) and \(\displaystyle \frac{X''}{X}+ 1= \alpha\). The boundary conditions are \(\displaystyle u(0, t)= X(0)T(t)= 0\) for all t, so we must have X(0)= 0, and \(\displaystyle u(1, t)= X(1)T(t)= 0\) for all t so we must have X(1)= 0.
So we start by solving the ordinary differential equation problem, \(\displaystyle X''+ (1- \alpha)X= 0\) with boundary conditions X(0)= X(1)= 0. But we don't know what \(\displaystyle \alpha\) is!
So we look at "cases". If \(\displaystyle \alpha= 1\), the equation is \(\displaystyle X''= 0\) and we have, by integrating twice, X(x)= Ax+ B. But then X(0)= B= 0 and X(1)= A+ B= 0 so the we must have X(x)= 0x+ 0= 0 for all x. That is the "trivial solution". There is no "non-trivia solution" so we cannot satisfy the condition that u(x, 0)= x.
If \(\displaystyle \alpha< 1\) then \(\displaystyle 1- \alpha\) is negative. We can make that explicit by writing \(\displaystyle 1- \alpha= -\lambda^2['tex] where \(\displaystyle \lambda\) can be any non-zero number (and, without loss of generality, we can take it to be positive). Now our equation is \(\displaystyle X''- \lambda^2 X= 0\). That is a "linear, homogenous, differential equation with constant coefficients". It "characteristic equation" is \(\displaystyle r^2- \lambda^2= 0\) which has roots \(\displaystyle r= \pm\lambda\) and so general solution \(\displaystyle X(x)= Ae^{\lambda x}+ Be^{-\lambda x}\). Now we have \(\displaystyle X(0)= A+ B= 0\) and \(\displaystyle X(1)= Ae^{\lambda}+ Be^{-\lambda}= 0\). From the first equation, B= -A, so the second equation becomes \(\displaystyle A(e^{\lambda}- e^{-\lambda})= 0\). Now \(\displaystyle e^{\lambda}\) is greater than 1 while \(\displaystyle e^{-\lambda}\) is less than 1. They are NOT equal so \(\displaystyle e^{\lambda}- e^{-\lambda}\) cannot be 0. That means we must have A= 0. But then B= 0 also so once again we have the trivial solution. There is no non-trivial solution in this case either.
So we must have \(\displaystyle \alpha> 1\) so that \(\displaystyle 1- \alpha> 0\). Write \(\displaystyle 1- \alpha= \lambda^2\) where, again, we can take \(\displaystyle \lambda\) to be positive. Now our equation is X''+ \lambda^2 X= 0. Now the "characteristic equation" is \(\displaystyle r^2+ \lambda^2= 0\) which has imaginary roots, \(\displaystyle r= \lambda i\) and \(\displaystyle r= -\lambda i\) and so general solution \(\displaystyle X(x)= A cos(\lambda x)+ B sin(\lambda x)\). \(\displaystyle X(0)= A= 0\). \(\displaystyle X(1)= B sin(\lambda)= 0\). We must have B= 0 or \(\displaystyle sin(\lambda)= 0\). We can have B non-zero, and so a non-trivial solution if and only if \(\displaystyle \lambda\) is a multiple of \(\displaystyle \pi\): \(\displaystyle \lambda= n\pi\). That is, we have a non-trivial solution to \(\displaystyle X''+ (1- \alpha)X= 0\) if and only if \(\displaystyle 1- \alpha= -n^2\pi^2\) so that \(\displaystyle \alpha= 1+ n^2\pi^2\), and in that case, we have \(\displaystyle X(x)= B sin(n\pi x)\).
Now, look at the equation for T. We had \(\displaystyle \frac{T'}{T}= \alpha= 1+ n^2\pi^2\) for n any integer. That is \(\displaystyle T'= (1+ n^2\pi^2)T\) which has general solution \(\displaystyle T= C e^{(1+ n^2\pi^2)t}\).
Putting that together with X, \(\displaystyle u(x, t)= Ce^{(1+ n^2\pi^2)t}sin(n\pi x)\).
To find a solution to the entire equation, since the "non-homogeneous part" is \(\displaystyle xe^{-t}\), look for a solution of the form \(\displaystyle u(x, t)= (ax+ b)e^{ct}\). Then \(\displaystyle u_x= ae^{ct}\), \(\displaystyle u_xx= 0\), and \(\displaystyle u_t= c(ax+ b)e^{ct}\). The equation becomes \(\displaystyle c(a+ b)e^{ct}= ((-ac- a)x- 2b))e^{ct}- xe^{-t}\).
Clearly, we must have \(\displaystyle c= -1\). Then we can divide by \(\displaystyle e^{ct}= e^{-t}\) and have \(\displaystyle -(a+ b)= \) or \(\displaystyle adx^2+ bdx+ dc= -ax^2+ (b- 1)x+ (2a+ c) for all x so we must have ad= -a, bd= b- 1, and dc= 2a+ c. d= -1, -b= b- 1 so b= 1/2, and -c= 2a+ c so a= c. That is, \(\displaystyle u(x, t)= Ce^{(1+ n^2\pi^2)t} sin(n\pi x)+
Now, we must have \(\displaystyle u(x, 0)= C sin(n\pi x)= x\). There is, of course, no "C" that will make that true for all x but, since this is a linear equation, we can form new solution as linear combinations of solutions. That is we can have solutions of the form \(\displaystyle u(x, t)= \sum_{n=0}^\infty C_n e^{(1+ n^2\pi^2)t} sin(n\pi x)\). That is a "Fourier sine series" solution. Now we have \(\displaystyle u(x, 0)= \sum_{n=0}^\infty C_n sin(n\pi x)= x\). That is, we must find the coefficients, \(\displaystyle C_n\), for the Fourier series equal to x and we have found the general solution to the associated homogeneous equation.\)\)\)
