Answer to Question #345687 in Differential Equations for Wesley

Question #345687

Find the general solution for the differential equation


-y"=6x + xe^x


(a) using method of undetermined coefficients


(b) using variation of parameters


1
Expert's answer
2022-05-31T07:00:29-0400

(a) method of undetermined coefficients

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

y=yu+yp1+yp2y=y_u+y_{p1}+y_{p2} ,

yu=0  λ1,2=0  yu=C1x+C2y_u\prime\prime=0\ \Longrightarrow\ \lambda_{1,2}=0\ \Longrightarrow\ y_u=C_1x+C_2 ,

yp1=6x; yp1=6xdx=3x2; yp1=3x2dx=x3y_{p1}\prime\prime=6x;\ \Longrightarrow y_{p1}\prime=\int6xdx=3x^2;\ \Longrightarrow y_{p1}=\int{3x^2dx}=x^3 ,

yp2=xexy_{p2}\prime\prime=xe^x ,

yp2(x)=(Ax+B)exy_{p2}\left(x\right)=\left(Ax+B\right)e^x ,

yp2(x)=Aex+ (Ax+B)ex=(A+Ax+B)exy_{p2}\prime\left(x\right)=Ae^x+\ \left(Ax+B\right)e^x=\left(A+Ax+B\right)e^x ,

yp2(x)=Aex+(A+Ax+B)ex=ex(2A+B+Ax)=xexy_{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=0A=1; 2A+B=0

A=1;B=2A=1; B=-2

yp2(x)=(Ax+B)ex=(x2)exy_{p2}\left(x\right)=\left(Ax+B\right)e^x=\left(x-2\right)e^x ,

Answer: y=C1x+C2+x3+(x2)exy=C_1x+C_2+x^3+\left(x-2\right)e^x .


(b) method of variation of parameters

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

y=yu+ypy=y_u+y_p ,

yu=0  λ1,2=0  yu=C1x+C2y_u\prime\prime=0\ \Longrightarrow\ \lambda_{1,2}=0\ \Longrightarrow\ y_u=C_1x+C_2 ,

make up a system of equations:

{C1y1+C2y2=0C1y1+C2y2=f(x)\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. ,

{C1x+C2=0C1+C20=6x+xex\left\{\begin{matrix}C_1\prime x+C_2\prime=0\\C_1\prime+C_2\prime0=6x+xe^x\\\end{matrix}\right. ,

{xC1+C2=0C1=6x+xex\left\{\begin{matrix}xC_1\prime +C_2\prime=0\\C_1\prime=6x+xe^x\\\end{matrix}\right. ,

C1(x)=(6x+xex)dx=3x2+ex(x1)+C1C_1\left(x\right)=\int\left(6x+xe^x\right)dx=3x^2+e^x\left(x-1\right)+C_1 ,

C2=(6x+xex)xdx=(6x2+x2ex)dx=2x3ex(x22x+2)+C2C_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=C1(x)x+C2(x)=(3x2+ex(x1)+C1)x2x3ex(x22x+2)+C2=x3+(x2)ex+C1x+C2y=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=x3+(x2)ex+C1x+C2y=x^3+\left(x-2\right)e^x+C_1x+C_2 .


NB: xexdx=(u=xdx=du)(dv=exdxv=ex)=xexexdx=ex(x1)+C\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 ,

x2exdx=(u=x22xdx=du)(dv=exdxv=ex)=x2ex2xexdx=x2ex2ex(x1)+C=ex(x22x+2)+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.



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