Question

In Problems 17-32 use the procedures developed in this chapter to find the general solution of each differential equation.

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

Short answer

The solution to the given differential equation is,

$y\left(x\right)=C_1{e}^x+C_{2}x{e}^x+\frac{1}{12}{x}^4{e}^x$

Step-by-step solution

Given Data

Differential equation is the equation which relate one or more unknown functions and their derivatives.

Given differential equation,

$y\text{'}\text{'}-2y\text{'}+y=x^2{e}^x$

Find auxiliary equation

Consider the following differential equation:

$y\text{'}\text{'}-2y\text{'}+y=x^2{e}^x$

First, find the complementary solution.

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

Let$y=e^{mx}$.
Then, the auxiliary equation is

$$\begin{gathered} m^2-2m+1=0 \\ {\left(m-1\right)}^2=0 \\ m=1,1 \end{gathered}$$

Therefore, the complementary solution is,

$y_c=c_1{e}^x+c_{2}x{e}^x$

Find particular solution

Now, find the particular solution.

$g\left(x\right)=x^2{e}^x$

Here, $x^2$suggests that we will have a quadratic form in $y_p$.

$$\begin{gathered} y_p=(A{x}^2+Bx+C)e^x \\ =A{x}^2{e}^x+Bx{e}^x+C{e}^x \end{gathered}$$

But this is duplicate because both $e^x$and$x{e}^x$belong to complementary solution as well.
If we multiply by $x$, then we get

$y_p=A{x}^3{e}^x+B{x}^2{e}^x+Cx{e}^x$

This is also duplicated, since$x{e}^x$is still presenting in complementary.
Now, again multiply with$x^2$to equation (1) to get the following:

$y_p=A{x}^4{e}^x+B{x}^3{e}^x+C{x}^2{e}^x$

This is the correct form as there is no duplication.

Final calculation

Therefore,

$$\begin{gathered} y_p\text{'}=4A{x}^3{e}^x+A{x}^4{e}^x+3B{x}^2{e}^x+B{x}^3{e}^x+2Cx{e}^x+2x^2{e}^x \\ =x{e}^x(A{x}^3+4A{x}^2+B{x}^2+3Bx+Cx+2C) \\ {y}_p\text{'}\text{'}=12A{x}^2{e}^x+8A{x}^3{e}^x+A{x}^4{e}^x+6Bx{e}^x+6B{x}^2{e}^x+B{x}^3{e}^x+2C{e}^x+4Cx{e}^x+C{x}^2{e}^x \\ =e^x(A{x}^4+8A{x}^3+B{x}^3+12A{x}^2+6B{x}^2+C{x}^2+6Bx+4Cx+2C) \end{gathered}$$

Now substitute the above values in the given differential equation.

localid="1664443843332" $e^x\left(A{x}^4+8A{x}^3+B{x}^3+12A{x}^2+6B{x}^2+C{x}^2+6Bx+4Cx+2C\right)-2(x{e}^x\left(A{x}^3+4A{x}^2+B{x}^2+3Bx+Cx+2C)\right)+A{x}^4{e}^x+B{x}^3{e}^3+C{x}^2{e}^x)]=x^2{e}^x$

By comparing the coefficients:

$$\begin{gathered} 12A=1\Rightarrow A=\frac{1}{12} \\ 6B=0\Rightarrow B=0 \\ 2C=0\Rightarrow C=0 \end{gathered}$$

Thus, the particular solution is$y_p=\frac{1}{12}{x}^4{e}^x$

Hence, the solution to the given differential equation is,

$$\begin{gathered} y=y_c+y_p \\ y\left(x\right)=C_1{e}^x+C_2{e}^x+\frac{1}{12}{x}^4{e}^x \end{gathered}$$