Given, $\dfrac{\partial U}{\partial t} = a_{pqr}f(t)-b_{pqr}U+\epsilon(V-U)-MU+S$
Prove $L(U)=\dfrac{a_{pqr}L(f(S))}{S_1-S_2}\cdot\left(\dfrac{S_1+\epsilon}{S-S_1}-\dfrac{S_2+\epsilon}{S-S_2}\right)$ where $S_1$ and $S_2$ are the roots of quadratic equation $S^2+(b_{pqr}+\epsilon+\gamma+M-S)+\epsilon b_{pqr}$
Following is my attempt:-
Applying laplace transform on both sides of the given equation
$L\left(\dfrac{\partial U}{\partial t}\right)= a_{pqr}L(f(t))-b_{pqr}L(U)+\epsilon L(V-U)-ML(U)+S\tag{1}$
$SL(U)=a_{pqr}+L(f(S))-b_{pqr}L(U)+\epsilon\left(L(V)-L(U)\right)-ML(U)+S\tag{2}$
In order to obtain U, I tried to take laplace inverse for equation $(2)$
$SL^{-1}(L(U))=a_{pqr}+L^{-1}(L(f(S))-b_{pqr}L^{-1}(L(U))+\epsilon(L^{-1}(L(V))-L^{-1}(L(U)))-ML^{-1}(L(U))+S\tag{3}$
$SU=a_{pqr}+f(S)-b_{pqr}U+\epsilon(V-U)-MU+S\tag{4}$
But I am not getting how to achieve the final proof from here. Please help me in this.