I've given the following PDE:
$$u_t=\dfrac{1}{2}u_{xx}-\dfrac{1}{10}u+1-x$$ with $$u(t,0)=0, \quad u_x(t,1)=0, \quad u(0,x)=20x(1-x).$$
I know how to use separation of variable methods and tried to use the substitution $u(t,x)=v(x)+w(t,x)$ but neither of them works.
Could someone give me some hints on how to solve this equation system analytically?
Given $$ u_t= a^2 \, u_{xx} - b^2 \, u + 1 - x $$ with $u(0, t)=0, \, u_x(1, t)=0, \, u(x, 0)= c \, x (1-x),$ then let $u(x, t) = w(x, t) + v(x)$ to obtain \begin{align} u_{t} &= w_{t} \\ u_{x} &= w_{x} + v_{x} \\ u_{x x} &= w_{x x} + v_{x x} \end{align} and $$ w_{t} = a^2 \, (w_{x x} + v_{x x} ) - b^2 \, (w + v) + 1 - x $$ or $$ w_{t} = a^2 \, w_{x x} - b^2 \, w + \left( a^2 \, v_{x x} - b^2 \, v + 1 - x \right).$$ Letting $$ a^2 \, v_{x x} - b^2 \, v + 1 - x = 0, $$ with $v(0) = 0$ and $v_{x}(1) = 0$ then $$ w_{t} = a^2 \, w_{x x} - b^2 \, w $$ with $ w(0, t) = 0, \, w_{x}(1, t) = 0, \, w(x, 0) = c \, x \, (1-x) - v(x).$ From here, $$ w(x, t) = e^{- \lambda \, t} \, F(x) $$ which leads to $$ F^{''} + \left(\frac{\lambda}{a^2} - b^2 \right) \, F = 0 $$ with $F(0) = 0$ and $F^{'}(1) = 0$. This gives $$ F(x) = B \, \sin\left(\frac{(2 n+1) \, \pi \, x}{2}\right) $$ with $$ \sqrt{\frac{\lambda_{n}}{a^2} - b^2} = \frac{(2 n + 1) \, \pi}{2} $$ or $$ \lambda_{n} = a^2 b^2 + \frac{a^2 \, (2 n +1)^2}{4}. $$ Now, $$ w(x, t) = \sum_{n=1}^{\infty} B_{n} \, e^{- \lambda_{n} \, t} \, \sin\left(\frac{(2 n+1) \, \pi \, x}{2}\right). $$
Returning to $v(x)$ the differential equation gives the solution $$ v(x) = A \, \cos\left(\frac{b x}{a} \right) + B \, \sin\left(\frac{b x}{a}\right) + \frac{1 - x}{b^2} $$ with the conditions $v(0) = 0$ and $v^{'}(1) = 0$. This gives $$ v(x) = \frac{1}{b^3} \, \left( b(1-x) - b \, \cos\left(\frac{b x}{a}\right) + (a \, \sec\left(\frac{b}{a}\right) + b \, \tan\left(\frac{b}{a}\right) \, \sin\left(\frac{b x}{a}\right) \right). $$
From here is it a matter of collecting the pieces and finding the Fourier Coefficients.