Question

In Problemsthe 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${\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$

${\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{}{\mathit{y}}_{\mathbf{1}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathit{l}\mathit{n}\mathbf{}\mathit{x}\mathbf{)}$

Short answer

$x^2\sin (\ln x)$

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,

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

Step 2: Find the second solution

We are given that,

$x^{2}y\text{'}\text{'}-3xy\text{'}+5y=0;y_1=x^2\cos (\ln x)$

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

$y\text{'}\text{'}-\frac{3}{x}y\text{'}+\frac{5}{x^2}y=0$

as the form,

$y\text{'}\text{'}+p\left(x\right)y\text{'}+q\left(x\right)y=0$, then we have $p\left(x\right)=-\frac{3}{x};q\left(x\right)=\frac{5}{x^2}$.

with a first solution $y_1\left(x\right)=x^2\cos (\ln x)$.

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

$$\begin{gathered} y_2\left(x\right)=y_1\left(x\right)\int \frac{e^{-\int p\left(x\right)dx}}{{\left(y_2\left(x\right)\right)}^2}dx \\ =x^2\cos (\ln x)\int \frac{e^{-\int -\frac{3}{x}dx}}{{\left(x^2\cos (\ln x)\right)}^2}dx \\ =x^2\cos (\ln x)\int \frac{e^{3\ln x}}{{\left(x^2\cos (\ln x)\right)}^2}dx \\ =x^2\cos (\ln x)\int \frac{x^3}{x^4{\cos }^2(\ln x)}dx \\ =x^2\cos (\ln x)\int \frac{1}{x{\cos }^2(\ln x)}dx \end{gathered}$$

Hence the value of $y_2\left(x\right)$is$x^2\cos (\ln x)\int \frac{1}{x{\cos }^2(\ln x)}dx$ .

Find the second solution of the given differential equation

To obtain the integration$\int \frac{1}{x^3{\cos }^2(\ln x)}dx$ , we have to use integration by substituting, where we use $u=\ln \text{x}$, then we have $du=\frac{1}{x}dx$, then we can obtain the integration as,

$$\begin{gathered} \int \frac{1}{x{\cos }^2(\ln x)}dx=\int \frac{1}{{\cos }^2\left(u\right)}du \\ =\int {\sec }^{2}udu \\ =\tan u \\ =\tan (\ln x) \end{gathered}$$

Now we can obtain the second solution of the given differential equation as,

$$\begin{gathered} y_2=x^2\cos (\ln x)\times \tan (\ln x) \\ =x^2\cos (\ln x)\times \frac{\sin (\ln x)}{\cos (\ln x)} \\ =x^2\sin (\ln x) \end{gathered}$$

Hence the second solution of the given differential equation is${\mathit{x}}^{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathit{l}\mathit{n}\mathbf{}\mathit{x}\mathbf{)}$ .