Answer to Question #210426 in Differential Equations for Bilal Ur Rehman

Question

In the following one solution of a second y1 order linear homogene DE is given. Find the second linearly independent solution y2 using the method of reduction of order.

2x²y" + 3xy'-y=0 ,. y1=(1/x)

(1-x²)y"-2xy'+2y=0 , y1=x

x²y"+2xy'-2y=0,. y1=x

x²y"+3xy'+y=0, y1=1/x

x²y"-x(x+2)y'=0,. y1=x

Answer 1

1.

$2x^2y'' + 3xy'-y=0 ,y_1=(1/x)$

$y(x)=u(x)y_1(x)$

$y(x)=u(x)x^{-1}$

$y'=u'x^{-1}-ux^{-2}$

$y''=u''x^{-1}-2u'x^{-2}+2ux^{-3}$

$2x^2(u''x^{-1}-2u'x^{-2}+2ux^{-3})+3x(u'x^{-1}-ux^{-2})$

$-ux^{-1}=0$

$2xu''-4u'+4ux^{-1}+3u'-3ux^{-1}-ux^{-1}=0$

$2xu''-u'=0$

$u'=w, u''=w'$

$2xw'-w=0$

$\dfrac{dw}{w}=\dfrac{dx}{2x}$

$\int\dfrac{dw}{w}=\int\dfrac{dx}{2x}$

$\ln|w|=\dfrac{1}{2}\ln|x|+\ln C_1$

$w=C_1\sqrt{x}$

$u'=C_1\sqrt{x}$

$\int du=\int C_1\sqrt{x}dx$

$u=C_2x^{3/2}+C_3$

$y(x)=u(x)x^{-1}$

$y(x)=C_2x^{1/2}+C_3x^{-1}$

2.

$(1-x^2)y''-2xy'+2y=0 , y_1=x$

$y(x)=u(x)y_1(x)$

$y(x)=u(x)x$

$y'=u'x+u$

$y''=u''x+2u'$

$(1-x^2)(u''x+2u')-2x(u'x+u)+2ux=0$

$x(1-x^2)u''+2u'-2x^2u'-2x^2u'-2ux+2ux=0$

$x(1-x^2)u''+(2-4x^2)u'=0$

$u'=w, u''=w'$

$x(1-x^2)w'+(2-4x^2)w=0$

$\dfrac{dw}{w}=\dfrac{2-4x^2}{x(1-x^2)}dx$

$\int\dfrac{dw}{w}=\int\dfrac{2-4x^2}{x(1-x^2)}dx$

$\dfrac{2-4x^2}{x(1-x^2)}dx=\dfrac{A}{x}dx+\dfrac{B}{1-x}dx+\dfrac{C}{1+x}dx$

$A(1-x^2)+Bx(1+x)+Cx(1-x)=2-4x^2$

$A-Ax^2+Bx+Bx^2+Cx-Cx^2=2-4x^2$

$A=2$

$B+C=0$

$-A+B-C=-4$

$A=2, B=-1, C=1$

$\int\dfrac{dw}{w}=\int\dfrac{2}{x}dx-\int\dfrac{1}{1-x}dx+\int\dfrac{1}{1+x}dx$

$\ln|w|=2\ln|x|+\ln|1-x|+\ln|1+x|+\ln C_1$

$w=C_1x^2(1-x^2)$

$u'=C_1x^2(1-x^2)$

$u=\dfrac{1}{3}C_1x^3-\dfrac{1}{5}C_1x^5+C_2$

$y=\dfrac{1}{3}C_1x^4-\dfrac{1}{5}C_1x^6+C_2x$

3.

$x^2y''+2xy-2y=0 , y_1=x$

$y(x)=u(x)y_1(x)$

$y(x)=u(x)x$

$y'=u'x+u$

$y''=u''x+2u'$

$x^2(u''x+2u')+2x(u'x+u)-2ux=0$

$x^3u''+2x^2u'+2ux-2ux=0$

$x^3u''+2x^2u'=0$

$u'=w, u''=w'$

$x^3w'+2x^2w=0$

If $x\not=0$

$\dfrac{dw}{w}=-\dfrac{2}{x}dx$

$\int\dfrac{dw}{w}=-\int\dfrac{2}{x}dx$

$\ln|w|=-2\ln|x|+\ln C_1$

$w=\dfrac{C_1}{x^2}$

$u'=\dfrac{C_1}{x^2}$

$u=-\dfrac{C_1}{x}+C_2$

$y=C_3+C_2x$

4.

$x^2y'' + 3xy'+y=0 ,y_1=(1/x)$

$y(x)=u(x)y_1(x)$

$y(x)=u(x)x^{-1}$

$y'=u'x^{-1}-ux^{-2}$

$y''=u''x^{-1}-2u'x^{-2}+2ux^{-3}$

$x^2(u''x^{-1}-2u'x^{-2}+2ux^{-3})+3x(u'x^{-1}-ux^{-2})$

$+ux^{-1}=0$

$xu''-2u'+2ux^{-1}+3u'-3ux^{-1}+ux^{-1}=0$

$xu''+u'=0$

$u'=w, u''=w'$

$xw'+w=0$

$\dfrac{dw}{w}=-\dfrac{dx}{x}$

$\int\dfrac{dw}{w}=-\int\dfrac{dx}{x}$

$\ln|w|=-\ln|x|+\ln C_1$

$w=\dfrac{C_1}{x}$

$u'=\dfrac{C_1}{x}$

$\int du=\int \dfrac{C_1}{x}dx$

$u=C_1\ln|x|+C_2$

$y(x)=u(x)x^{-1}$

$y(x)=\dfrac{C_1\ln x}{x}+\dfrac{C_2}{x}$

5.

$x^2y''-x(x+2)y'=0 , y_1=x$

If $x\not=0$

$xy''-(x+2)y'=0$

$y'=w, y''=w'$

$xw'-(x+2)w=0$

If $x\not=0$

$\dfrac{dw}{w}=\dfrac{x+2}{x}dx$

$\int\dfrac{dw}{w}=\int\dfrac{x+2}{x}dx$

$\ln |w|=x+\ln x^2+\ln C_1$

$y'=C_1x^2e^x$

$y=\int C_1x^2e^xdx$ $\int C_1x^2e^xdx=C_1x^2e^x-2C_1\int xe^xdx$

$=C_1x^2e^x-2C_1xe^x+2C_1\int e^xdx$

$=C_1x^2e^x-2C_1xe^x+2C_1e^x+C_2$

$y=C_1x^2e^x-2C_1xe^x+2C_1e^x+C_2$