Question

In Problems $\mathbf{1}\mathbf{-}\mathbf{16}$ 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${\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$

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

Short answer

The second solution of the given differential equation is $-e^{-x}$

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$$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

$y\text{'}\text{'}+2y\text{'}+y=0$

as the form

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

with a first solution$y_1\left(x\right)=x{e}^{-x}$ , and we have to obtain the second solution as

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

Therefore, the second solution of the given differential equation is $\mathbf{-}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$