Answer to Question #218040 in Differential Equations for norman

Question

solve the following question

1. dy/dx+3y=3x²e^-x

Answer 1

Let us solve the following differential equation $\frac{dy}{dx}+3y=3x^2e^{-x}.$ For this let us multply both parts by $e^{3x}:$ $e^{3x}\frac{dy}{dx}+3e^{3x}y=3x^2e^{-x}e^{3x}.$ It follows that $(e^{3x}y)'=3x^2e^{2x}.$ Then $e^{3x}y=3\int x^2e^{2x}dx=|u=x^2,dv=e^{2x}dx,du=2xdx, v=\frac{1}{2}e^{2x}|=\frac{3}{2}x^2e^{2x}-3\int xe^{2x}dx=|u=x,dv=e^{2x},du=dx,v=\frac{1}{2}e^{2x}|=\frac{3}{2}x^2e^{2x}-\frac{3}{2}xe^{2x}+\frac{3}{2}\int e^{2x}dx=\frac{3}{2}x^2e^{2x}-\frac{3}{2}xe^{2x}+\frac{3}{4}e^{2x}+C.$

It follows that the general solution is $y=e^{-3x}(\frac{3}{2}x^2e^{2x}-\frac{3}{2}xe^{2x}+\frac{3}{4}e^{2x}+C)=\frac{3}{2}x^2e^{-x}-\frac{3}{2}xe^{-x}+\frac{3}{4}e^{-x}+Ce^{-3x} .$