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

$\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{}\mathbf{}\mathbf{}{\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathit{l}\mathit{n}\mathbf{}\mathit{x}$

Short answer

The second solution of the given differential equation is $y_2=1$

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$$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

$xy\text{'}\text{'}+y\text{'}=0$

which is

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

as the form

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

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

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

Obtain the integration method

To obtain the integration $\int \frac{1}{x{\ln }^{2}x}dx$, 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

$$\begin{gathered} \int \frac{1}{x{\ln }^{2}x}dx=\int \frac{1}{u^2}du \\ =-\frac{1}{u} \\ =-\frac{1}{\ln x} \end{gathered}$$

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

$$\begin{gathered} y_2=\ln x\times -\frac{1}{\ln x} \\ =-1 \\ =c_2\times 1 \end{gathered}$$

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

Therefore, the second solution of the given differential equation is ${\mathit{y}}_{\mathbf{2}}\mathbf{=}\mathbf{1}$