Question

For Problems 1 to 4, find one (simple) solution of each differential equation by series, and then find the second solution by the "reduction of order" method, Chapter 8, Section 7 (e).

$\mathbf{(}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{)}\mathit{y}\mathbf{"}\mathbf{-}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The answer is stated below.

Step-by-step solution

Concept of reduction of order method

Reduction of order is a technique in mathematics of solving second-order linear ordinary differential equations. It is employed when one solution y₁(x)is known and the second linearly independent solution y2(x) is desired. The method also applies to n-th order equations. In this case (n-1) the ansatz will yield an order equation for v.

Use the concept of the reduction order for calculation

$$" width="9">" width="9">x2+1y"-xy'+y=0.

Find the simple solution and then after the second solution by the reduction of order the differential equation.

Can be written as follows:

$y"+\left(-\frac{x}{x^2+1}\right)y\text{'}+\left(\frac{1}{x^2+1}\right)y=0$

Comparing the differential equation as follows:

With the equation, $y"+P\left(x\right)y\text{'}+Q\left(x\right)y=0$.

Now, $$" width="9">

localid="1664376578077" $$\begin{gathered} P\left(x\right)=-\frac{x}{x^2+1}andQ\left(x\right)=\frac{1}{x^2+1} \\ AfteranalyzingP\left(x\right)andQ\left(x\right);x^2+1=0 \\ Therefore,x=\pm i \\ so,thesimplesolutionofthisdiferentequationconcentrayedat0anditwillconverge \\ onlyfor\left|x\right|<1andhere1isthedis\tan ceinthecomplexplanefrom0toeitherior-i \\ now,assume,y=\sum _{n=0}^{\infty }{c}_n{x}^{n}isthesolutionofthegivendifferentialequation\left(1\right). \\ so,y\text{'}=\sum _{n=0}^{\infty }n{c}_n{X}^{n-1}andy"=\sum _{n=0}^{\infty }n(n-1)c_n{X}^{n-2} \\ now,get \end{gathered}$$

$$\begin{gathered} \left(x^2+1\right)y"-xy\text{'}+y \\ =\left(x^2+1\right)\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}-x\sum _{n=1}^{\infty }n{c}_n{x}^{n-1}+\sum _{n=0}^{\infty }{c}_n{x}^n \\ =\sum _{n=2}^{\infty }n(n-1)c_n{x}^n+\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}-x\sum _{n=1}^{\infty }n{c}_n{x}^n+\sum _{n=0}^{\infty }{c}_n{x}^n \\ =2c_2{x}^0+c_0{x}^0+6c_{3}x-c_{1}x \end{gathered}$$