Not sure how to even attempt this question, can anyone offer any advice on where to begin?
Find solutions v = v(x,y) of the PDE i) vy+ v = e^xy
Since there is no derivative with respect to x in this equation, you can treat x as a constant so this becomes an
ordinary differential equation.
The "associated homogeneous equation" is \(\displaystyle v'+ v= 0\) which has "characteristic equation" r+ 1= 0 so r= -1. The general solution is \(\displaystyle v(y)= Ce^{-y}\). We look for a solution to the entire equation of the form \(\displaystyle v= Ae^{xy}\) where, again, we treat x as a constant. Then \(\displaystyle v'= Axe^{xy}\) and the equation becomes \(\displaystyle Axe^{xy}+ Ae^{xy}= A(x+ 1)e^{xy}=e^{xy}\). We must have \(\displaystyle A= \frac{1}{x+ 1}\). Notice that if x= -1 that does not exist. That is because \(\displaystyle e^{-y}\) is already a solution to the homogeneous equation.
In the case that x= -1 we look for a solution of the form \(\displaystyle v= Aye^{-y}\) so that \(\displaystyle v'= Ae^{-y}- Aye^{-y}\) and the equation becomes \(\displaystyle Ae^{-y}- Aye^{-y}+ Aye^{-y}= Ae^{-y}= e^{-y}\) so A= 1.
The general solution to the equation \(\displaystyle v_y+ v= e^{xy}\) is
\(\displaystyle v(x, y)= f(x)e^{-y}+ \frac{e^{-y}}{x+ 1}\) if x is not -1
\(\displaystyle v(-1, y)= Ce^{-y}+ ye^{-y}\) if x= -1.
Notice that in the case that x is not -1 the undetermined constant, "C", is replaced by the undetermined function, "f(x)". That is because, since we are treating x as a constant, the "constant" C might in fact depend on x. We don't do that in the case that x= -1 because then C= f(-1).
