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

$\mathbf{6}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{}\mathbf{}\mathbf{}{\mathit{y}}_{\mathbf{1}}\mathbf{=}{\mathit{e}}^{\mathbf{x}\mathbf{/}\mathbf{3}}$

Short answer

The second solution of the given differential equation is$e^{-\frac{1}{2}x}$

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

$6y\text{'}\text{'}+y\text{'}-y=0$

which is

$y\text{'}\text{'}+\frac{1}{6}y\text{'}-\frac{1}{6}y=0$

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)=\frac{1}{6}\text{and}q\left(x\right)=-\frac{1}{6}$$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}{6}\text{and}q\left(x\right)=-\frac{1}{6}$

with a first solution $y_1\left(x\right)=e^{\frac{1}{3}x}$, and we have to obtain the second solution sing the formula of reduction of order as

$$\begin{gathered} y_2\left(x\right)=y_1\left(x\right)\int \frac{e^{-\int p\left(x\right)dx}}{{\left(y_1\right)}^2}dx \\ =e^{\frac{1}{3}x}\int \frac{e^{-\frac{1}{6}x}}{{\left(e^{\frac{1}{3}x}\right)}^2}dx \\ =e^{\frac{1}{3}x}\int \frac{e^{-\frac{1}{6}x}}{e^{\frac{2}{3}x}}dx \\ =e^{\frac{1}{3}x}\int e^{-\frac{1}{6}x}\times e^{-\frac{2}{3}x}dx \\ =e^{\frac{1}{3}x}\int e^{-\frac{5}{6}x}dx \\ =-\frac{6}{5}{e}^{\frac{1}{3}x}\times e^{-\frac{5}{6}x} \\ =-\frac{6}{5}{e}^{-\frac{1}{2}x} \\ =c_2{e}^{-\frac{1}{2}x} \end{gathered}$$

Therefore, the second solution of the given differential equation$6y\text{'}\text{'}+y\text{'}-y=0$, is $c_2{e}^{-\frac{1}{2}x}$where $c_2=-\frac{6}{5}$and we let it equal 1.

Therefore, the second solution of the given differential equation is ${\mathit{e}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{x}}$