Question

In Problems21 and 22 the indicated function${\mathit{y}}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$is a solution of the given differential equation. Use formula (5),to find a second solution${\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$expressed in terms of an integral-defined function. See(iii) in the Remarks.

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

Short answer

$y_2\left(x\right)=e^x{\int }_{x_0}^x\frac{\sqrt{t}}{e^t}dt$

Step-by-step solution

Step 1:Definereduction of order

To solve a second order differential equation, the reduction of order approach is based on the principle of solving first order differential equations one after the other that have been generated from the original second order equation to simplify the problem.

The general formula is,

$y_2\left(x\right)=y_1\left(x\right)\int \frac{e^{-\int p\left(x\right)dx}}{{\left(y_2\left(x\right)\right)}^2}dx.$

Find the value of y2(x)

We are given that,

$2xy\text{'}\text{'}-(2x+1)y\text{'}+y=0;y_1=e^x$

This is a second order differential equation. It can be written as,

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

as the form,

.$y\text{'}\text{'}+p\left(x\right)y\text{'}+q\left(x\right)y=0$

with a first solution $y_1=e^x$$y_1=e^x$.

We have to obtain the second solution using the formula for reduction of order as,

$$\begin{gathered} y_2\left(x\right)=y_1\left(x\right)\int \frac{e^{-\int p\left(x\right)dx}}{{\left(y_2\left(x\right)\right)}^2}dx. \\ =e^x{\int }_{x_0}^x\frac{e^{\int \left(1+\frac{1}{2t}\right)dt}}{e^{2t}}dt \\ =e^x{\int }_{x_0}^x\frac{e^{\left(t+\frac{\ln t}{2}\right)}}{e^{2t}}dt \\ =e^x{\int }_{x_0}^x\frac{e^t{e}^{\ln \sqrt{t}}}{e^{2t}}dt \\ =e^x{\int }_{x_0}^x\frac{\sqrt{t}}{e^t}dt \end{gathered}$$

Hence the second solution of the given differential equation is.${\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{e}}^{\mathbf{x}}{\mathbf{\int }}_{{\mathbf{x}}_{\mathbf{0}}}^{\mathbf{x}}\frac{\sqrt{\mathbf{t}}}{{\mathbf{e}}^{\mathbf{t}}}\mathit{d}\mathit{t}$