I want to solve the "modified" heat equation $$ \frac{\partial y}{\partial t}=a\frac{\partial^2 y}{\partial x^2} +b\frac{\partial y}{\partial x} +cy+d $$
I assumed that a, b, c and d are all constant
I tried using Laplace transform. $$ L(y(x,t))=\int_{0}^{\infty}e^{-st}u(x,t)dt=U(x,s) $$ $$ L(\frac{\partial y(x,t)}{\partial t})=sU(x,s)-u(x,0) $$ $$ L(\frac{\partial^2 y(x,t)}{\partial x^2})=\frac{\partial^2 U(x,s)}{\partial x^2} =U_{xx}(x,s) $$ $$ L(\frac{\partial y(x,t)}{\partial x})=\frac{\partial U(x,s)}{\partial x} =U_{x}(x,s) $$ Now equation looks like $$ sU(x,s)-u(x,0) =aU_{xx}(x,s)+bU_{x}(x,s)+cU(x,s)+d $$ or defining $f(x)=-d-u(x,0)$ $$ aU_{xx}(x,s)+bU_{x}(x,s)+(c-s)U(x,s)=f(x) $$ Now, homogeneous solution is $$ U_h(x,s)=U_1+U_2=c_1e^{\lambda_1x}+c_2e^{\lambda_2x} $$ where $\lambda_1=\frac{-b+\sqrt{b^2-4a(c-s)}}{2a}$, $\lambda_2=\frac{-b-\sqrt{b^2-4a(c-s)}}{2a}$,
and for particular solution, I used variation of parameters $$ U_p=-U_1\int\frac{U_2}{U_1U'_2-U_2U'_1}f(x)dx+U_2\int\frac{U_1}{U_1U'_2-U_2U'_1}f(x)dx $$
Then something hit me. It is wise to see if other people have already done this.
So, is there any literature that deals with this partial differential equation with the general solution?
(Anyway,does this have solution? just like integral of heat kernel in usual "heat equation"?)