Question

In Problemsthe indicated function${\mathit{y}}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$is a solution of the associated homogeneous equation. Use the method of reduction of order to find a second solution${\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$of the homogeneous equation and a particular solution${\mathit{y}}_{\mathit{P}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$of the given nonhomogeneous equation.

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

Short answer

$y_2\left(x\right)=e^{3x}$and$y_p=\frac{1}{3}x+\frac{4}{9}$ .

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,

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

Find the value of y2(x)

We are given that,

$y\text{'}\text{'}-4y\text{'}+3y=x;y_1=e^x$

This is a second order nonhomogeneous differential equation.

The homogeneous equation is

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

as the form,

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

with a first solution$y_1=e^x$ .

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

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

Hence the value is ${\mathit{e}}^{\mathbf{3}\mathbf{x}}$

Find the second solution of the given differential equation

To find the particular solution for the nonhomogeneous equation, we have to assume that $y=Ax+B$.

We have to differentiate our assumption with respect to as,

$$\begin{gathered} y\text{'}=A······\text{(1)} \\ y\text{'}\text{'}=0·····\text{(2)} \end{gathered}$$

Substitute and in the given differential equation,

$$\begin{gathered} 0-4A+3Ax+3B=x \\ 3Ax+(3B-4A)=1 \\ 3A=1then:A=\frac{1}{3} \\ 3B-4A=0then:B=\frac{4}{9} \end{gathered}$$

Then we can obtain,$y_p=\frac{1}{3}x+\frac{4}{9}$

Hence the particular solution of the given nonhomogeneous equation is${\mathit{y}}_{\mathbf{p}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathit{x}\mathbf{+}\frac{\mathbf{4}}{\mathbf{9}}$