Question

In Problems the indicated function${\mathit{y}}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ is a solution of the given differential equation. Use reduction of order or formula (5), as instructed, to find a second solution${\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$

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

Short answer

The second solution of the given differential equation is $y_2=x\cos (\ln x)$

Step-by-step solution

Definition of Reduction of order

• Reduction of order is a method for solving second-order linear ordinary differential equations in mathematics.
• When one solution is known and a second linearly independent solution is sought, this method is used.
• The approach can be used to solve n-th order equations as well.
• $y_2=y_1\left(x\right)\int \frac{e^{-\int P\left(x\right)dx}}{y_1^2\left(x\right)}dx$

Find the second solution

The second order differential equation

$x^{2}y\text{'}\text{'}-xy\text{'}+2y=0$

which is

$y\text{'}\text{'}-\frac{1}{x}y\text{'}+\frac{2}{x^2}y=0$

as the form

$y\text{'}\text{'}+p\left(x\right)y\text{'}+q\left(x\right)y=0,\text{hen we have}p\left(x\right)=-\frac{1}{x}\text{and}q\left(x\right)=\frac{2}{x^2},$

with a first solution$y_1\left(x\right)=x\sin (\ln x)$ , and we have to obtain the second solution using the formula of reduction of order as

$$\begin{gathered} y_2\left(x\right)=y_1\left(x\right)\int \frac{e^{-\int p\left(x\right)dx}}{{\left(y_2\left(x\right)\right)}^2}dx \\ =x\sin (\ln x)\int \frac{e^{-\int -\frac{1}{x}dx}}{{(x\sin (\ln x\left)\right)}^2}dx \\ =x\sin (\ln x)\int \frac{e^{\ln x}}{{(x\sin (\ln x\left)\right)}^2}dx \\ =x\sin (\ln x)\int \frac{x}{x^2{\sin }^2(\ln x)}dx \\ =x\sin (\ln x)\int \frac{1}{x{\sin }^2(\ln x)}dx \end{gathered}$$

Obtain the integration method

To obtain the integration $\int \frac{1}{x{\sin }^2(\ln x)}dx$, we have to use integration by substituting, where we use$u=\ln x$ , then we have$du=\frac{1}{x}dx$ , then we can obtain the integration as

$$\begin{gathered} \int \frac{1}{x{\sin }^2(\ln x)}dx=\int \frac{1}{{\sin }^2\left(u\right)}du \\ =\int {\csc }^{2}udu \\ =-\cot u \\ =-\cot (\ln x) \end{gathered}$$

Then we can obtain the second solution of the given D.E as

$$\begin{gathered} y_2=x\sin (\ln x)\times -\cot (\ln x) \\ =x\sin (\ln x)\times -\frac{\cos (\ln x)}{\sin (\ln x)} \\ =-x\cos (\ln x) \\ =c_{2}x\cos (\ln x) \\ =x\cos (\ln x) \end{gathered}$$

Therefore, the second solution of the given differential equation$x^{2}y\text{'}\text{'}-xy\text{'}+2y=0$, where we have $c_2=-1$ is $y_2=x\cos (\ln x)$

Therefore, the second solution of the given differential equation is${\mathit{y}}_{\mathbf{2}}\mathbf{=}\mathit{x}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathit{l}\mathit{n}\mathbf{}\mathit{x}\mathbf{)}$