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

$\mathbf{9}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{12}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{}\mathbf{}\mathbf{}{\mathit{y}}_{\mathbf{1}}\mathbf{=}{\mathit{e}}^{\mathbf{2}\mathbf{x}\mathbf{/}\mathbf{3}}$

Short answer

The second solution of the given differential equation is $x{e}^{\frac{2x}{3}}$

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

$9y\text{'}\text{'}-12y\text{'}+4y=0$

which is

$y\text{'}\text{'}-\frac{4}{3}y\text{'}+\frac{4}{9}y=0$

as the form

$y\text{'}\text{'}+p\left(x\right)y\text{'}+q\left(x\right)y=0,\text{then we have}p\left(x\right)=-\frac{4}{3}\text{and}q\left(x\right)=\frac{4}{9},$

with a first solution $y_1\left(x\right)=e^{\frac{2}{3}x}$, and we have to obtain the second solution the formula of reduction of order as

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

Therefore, the second solution of the given differential equation is$\mathit{x}{\mathit{e}}^{\frac{\mathbf{2}\mathbf{x}}{\mathbf{3}}}$