Solve the following PDE $$\partial_{yy} u + \partial_y u + 3u = 4e^{3x}, \quad u = u(x, y).$$
What I get is the following but when verified it wasn't what I expected to be and didn't get the solution I wanted
can someone verify
$$u=v(x)e^{-3}\left(k\cos(2y)+w\sin(2y)+\frac1{10}e^3y\right)$$
$$u''+6u'+13u=4e^{3y}$$ The characteristic equation is $$r^2+6r+13=0 \implies \Delta =36-4.13=-16$$ $$\implies S_r=\{-3 \pm 2i\}$$ The solution of the homogeneous equation is $$u(x,y)=e^{-3y}(c_1(x)\cos (2y)+c_2(x) \sin (2y))$$ For the particular solution $$u_p=Ae^{3y} \implies 9A+18A+13A=4 $$ $$\implies A=\frac 1 {10}$$ Finally: $$u(x,y)=e^{-3y}(c_1(x)\cos (2y)+c_2(x) \sin (2y))+\frac 1 {10}e^{3y}$$