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.

1. 2x^²y" + 3xy'-y=0 ,.
2. (1-x²)y"-2xy'+2y=0 ,
3. x²y"+2xy'-2y=0,
4. x²y"+3xy'+y=0,
5. x²y"-x(x+2)y'=0,

Answer 1

1.

Euler Equation.

$y=x^m$

Then:

$x^2m(m-1)x^{m-2}+3/2xmx^{m-1}-x^m/2=0$

Characteristic Equation:

$m(m-1)+\frac{3}{2}m+\frac{1}{2}=0$

or

$m^2+\frac{1}{2}m+\frac{1}{2}=0$

$m_{1,2}=\frac{-1\pm i\sqrt{3}}{4}$

Thus, the general solution is

$y=x^{-1/4}[c_1cos(\sqrt{3}ln|x|/4)+c_2sin(\sqrt{3}ln|x|/4)]$

2.

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

$y_1=x$

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

$y_2=\int\frac{e^{-\int p(x)dx}}{y^2_1}dx$

where p(x)=-2x/(1+x²)

Then:

$\int \frac{2x}{1+x^2}dx=ln|1+x^2|$

$y_2=\int \frac{dx}{x^2(1+x^2)}=-1/x+arctanx$

$\frac{1}{x^2(1+x^2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{1+x^2}$

$Ax(1+x^2)+B(1+x^2)+Cx^3+Dx^2=1$

$A+C=0$

$B+D=0$

$A=0,B=1,D=-1,C=0$

3.

x²y"+2xy'-2y=0

$y=x^m$

The characteristic equation:

$m(m-1)+2m-2=0$

$m_{1}=\frac{-1+ 3}{2}=1,m_2=-2$

The general solution is

$y=c_1x+c_2x^{-2}$

4.

x²y"+3xy'+y=0

$y=x^m$

The characteristic equation:

$m(m-1)+3m+1=0$

$m_{1,2}=\frac{-2}{2}=-1$

$y_1=x^{-1},y_2=x^{-1}ln|x|$

5.

x²y"-x(x+2)y'=0

$y'=z,y''=dz/dx$

Then:

$x^2z'-x(x+2)z=0$

$\frac{dz}{z}=\frac{x(x+2)}{x^2}dx$

$ln|z|=x+2ln|x|+c_1$

$z=x^2e^{c_1x}$

$y=\intop x^2e^{c_1x}dx=\frac{e^{c_1x}(c_1^2x^2-2c_1x+2)}{c_1^3}+c_2$