Answer to Question #345687 in Differential Equations for Wesley

Question

Find the general solution for the differential equation

-y"=6x + xe^x

(a) using method of undetermined coefficients

(b) using variation of parameters

Answer 1

(a) method of undetermined coefficients

$y\prime\prime=6x+xe^x$

$y=y_u+y_{p1}+y_{p2}$ ,

$y_u\prime\prime=0\ \Longrightarrow\ \lambda_{1,2}=0\ \Longrightarrow\ y_u=C_1x+C_2$ ,

$y_{p1}\prime\prime=6x;\ \Longrightarrow y_{p1}\prime=\int6xdx=3x^2;\ \Longrightarrow y_{p1}=\int{3x^2dx}=x^3$ ,

$y_{p2}\prime\prime=xe^x$ ,

$y_{p2}\left(x\right)=\left(Ax+B\right)e^x$ ,

$y_{p2}\prime\left(x\right)=Ae^x+\ \left(Ax+B\right)e^x=\left(A+Ax+B\right)e^x$ ,

$y_{p2}\prime\prime\left(x\right)=Ae^x+\left(A+Ax+B\right)e^x=e^x\left(2A+B+Ax\right)=xe^x$,

$A=1; 2A+B=0$

$A=1; B=-2$

$y_{p2}\left(x\right)=\left(Ax+B\right)e^x=\left(x-2\right)e^x$ ,

Answer: $y=C_1x+C_2+x^3+\left(x-2\right)e^x$ .

(b) method of variation of parameters

$y\prime\prime=6x+xe^x=f\left(x\right)$ ,

$y=y_u+y_p$ ,

$y_u\prime\prime=0\ \Longrightarrow\ \lambda_{1,2}=0\ \Longrightarrow\ y_u=C_1x+C_2$ ,

make up a system of equations:

$\left\{\begin{matrix}C_1\prime y_1+C_2\prime y_2=0\\C_1\prime y_1\prime+C_2\prime y_2\prime=f\left(x\right)\\\end{matrix}\right.$ ,

$\left\{\begin{matrix}C_1\prime x+C_2\prime=0\\C_1\prime+C_2\prime0=6x+xe^x\\\end{matrix}\right.$ ,

$\left\{\begin{matrix}xC_1\prime +C_2\prime=0\\C_1\prime=6x+xe^x\\\end{matrix}\right.$ ,

$C_1\left(x\right)=\int\left(6x+xe^x\right)dx=3x^2+e^x\left(x-1\right)+C_1$ ,

$C_2=-\int\left(6x+xe^x\right)xdx=-\int\left(6x^2+x^2e^x\right)dx=-2x^3-e^x\left(x^2-2x+2\right)+C_2$ ,

$y=C_1\left(x\right)x+C_2\left(x\right)=\left(3x^2+e^x\left(x-1\right)+C_1\right)x-2x^3-e^x\left(x^2-2x+2\right)+C_2=x^3+\left(x-2\right)e^x+C_1x+C_2$

Answer: $y=x^3+\left(x-2\right)e^x+C_1x+C_2$ .

NB: $\int{xe^xdx}=\binom{u=x}{dx=du}\binom{dv=e^xdx}{v=e^x}=xe^x-\int{e^xdx}=e^x\left(x-1\right)+C$ ,

$\int{x^2e^xdx}=\binom{u=x^2}{2xdx=du}\binom{dv=e^xdx}{v=e^x}=x^2e^x-2\int{xe^xdx}=x^2e^x-2e^x\left(x-1\right)+C=e^x\left(x^2-2x+2\right)+C.$