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 $x\not=0$

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

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

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

$$y'=C_1x^2e^x$$

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

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

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

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