Answer to Question #222706 in Differential Equations for sam

Question #222706

x2y''+xy'+4y=2xlnx


1
Expert's answer
2021-08-11T19:07:46-0400
x2y+xy+4y=2xlnxx^2y''+xy'+4y=2x\ln x

Homogeneous equation


x2y+xy+4y=0x^2y''+xy'+4y=0


Let y=xry=x^r


y=rxr1y'=rx^{r-1}


y=r(r1)xr2y''=r(r-1)x^{r-2}


r(r1)+r+4=0r(r-1)+r+4=0

r=±2ir=\pm2 i

The general solution of the homogeneous equation is


yh=C1cos(2lnx)+C2sin(2lnx)y_h=C_1\cos(2\ln x)+C_2\sin(2\ln x)


x2y+xy+4y=2xlnxx^2y''+xy'+4y=2x\ln x

Divide both sides by x2x^2


y+1xy+4x2y=2xlnx=>g(x)=2xlnxy''+\dfrac{1}{x}y'+\dfrac{4}{x^2}y=\dfrac{2}{x}\ln x=>g(x)=\dfrac{2}{x}\ln x

Wronskian


W(y1,y2)=cos(2lnx)sin(2lnx)2xsin(2lnx)2xcos(2lnx)=2xW(y_1, y_2)=\begin{vmatrix} \cos(2\ln x)& \sin(2\ln x) \\ - \dfrac{2}{x}\sin (2\ln x) & \dfrac{2}{x}\cos(2\ln x) \end{vmatrix}=\dfrac{2}{x}

W1=0sin(2lnx)2xlnx2xcos(2lnx)=2xlnxsin(2lnx)W_1=\begin{vmatrix} 0& \sin(2\ln x) \\ \dfrac{2}{x}\ln x & \dfrac{2}{x}\cos(2\ln x) \end{vmatrix}=-\dfrac{2}{x}\ln x \sin(2\ln x)


W2=cos(2lnx)02xsin(2lnx)2xlnx=2xlnxcos(2lnx)W_2=\begin{vmatrix} \cos(2\ln x) & 0 \\ - \dfrac{2}{x}\sin (2\ln x) & \dfrac{2}{x}\ln x \end{vmatrix}=\dfrac{2}{x}\ln x \cos(2\ln x)

u1=W1Wdx=2xlnxsin(2lnx)2xdxu_1=\int\dfrac{W_1}{W}dx=\int\dfrac{-\dfrac{2}{x}\ln x \sin(2\ln x)}{\dfrac{2}{x}}dx

=lnxsin(2lnx)dx=-\int \ln x \sin(2\ln x)dx

=125x(5lnx+3)sin(2lnx)+125(10lnx4)cos(2lnx)=-\dfrac{1}{25}x(5\ln x+3)\sin(2\ln x)+\dfrac{1}{25}(10\ln x-4)\cos(2\ln x)


u2=W2Wdx=2xlnxcos(2lnx)2xdxu_2=\int\dfrac{W_2}{W}dx=\int\dfrac{\dfrac{2}{x}\ln x \cos(2\ln x)}{\dfrac{2}{x}}dx

=lnxcos(2lnx)dx=\int \ln x \cos(2\ln x)dx

=125x(10lnx4)sin(2lnx)+125(5lnx+3)cos(2lnx)=\dfrac{1}{25}x(10\ln x-4)\sin(2\ln x)+\dfrac{1}{25}(5\ln x+3)\cos(2\ln x)

Particular solution of the non-homogeneous equation is


yp=u1y1+u2y2y_p=u_1y_1+u_2y_2

yp=125x(5lnx+3)sin(2lnx)cos(2lnx)y_p=-\dfrac{1}{25}x(5\ln x+3)\sin(2\ln x)\cos(2\ln x)

+125(10lnx4)cos2(2lnx)+\dfrac{1}{25}(10\ln x-4)\cos^2(2\ln x)

+125x(10lnx4)sin2(2lnx)+\dfrac{1}{25}x(10\ln x-4)\sin^2(2\ln x)

+125(5lnx+3)sin(2lnx)cos(2lnx)+\dfrac{1}{25}(5\ln x+3)\sin(2\ln x)\cos(2\ln x)

The general solution of the homogeneous equation is


y=C1cos(2lnx)+C2sin(2lnx)y=C_1\cos(2\ln x)+C_2\sin(2\ln x)

125x(5lnx+3)sin(2lnx)cos(2lnx)-\dfrac{1}{25}x(5\ln x+3)\sin(2\ln x)\cos(2\ln x)

+125(10lnx4)cos2(2lnx)+\dfrac{1}{25}(10\ln x-4)\cos^2(2\ln x)

+125x(10lnx4)sin2(2lnx)+\dfrac{1}{25}x(10\ln x-4)\sin^2(2\ln x)

+125(5lnx+3)sin(2lnx)cos(2lnx)+\dfrac{1}{25}(5\ln x+3)\sin(2\ln x)\cos(2\ln x)




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