Question

In Problems 1–26 solve the given differential equation by undetermined coefficients.

Short answer

$y=c_1{e}^{\frac{1}{2}x}+c_{2}x{e}^{\frac{1}{2}x}+\frac{1}{2}{x}^2{e}^{\frac{1}{2}x}+12$

Step-by-step solution

Finding the corresponding homogeneous equation

We are given with the non-homogeneous differential equation $y\text{'}\text{'}-y\text{'}+\frac{1}{4}y=3+e^{\frac{x}{2}}$and we need to find its general solution by following the below method:

Firstly, we need to find the solution for the corresponding homogeneous equation,

$y\text{'}\text{'}-y\text{'}+\frac{1}{4}y=0$

Consider $y=e^{mx}$as the solution of the differential equation,

Substitute,

$y=e^{mx},y\text{'}=m{e}^{mx}$,$y\text{'}\text{'}=m^2{e}^{mx}$into $y\text{'}\text{'}-y\text{'}+\frac{1}{4}y=0$to obtain the auxiliary equation.

$$\begin{gathered} m^2{e}^{mx}-m{e}^{mx}+\frac{1}{4}{e}^{mx}=0 \\ {e}^{mx}\left(m^2-m+\frac{1}{4}\right)=0 \end{gathered}$$

Since $e^{mx}$cannot be equal to , then we have

role="math" localid="1663837529063" $$\begin{gathered} m^2-m+\frac{1}{4}=0 \\ {\left(m-\frac{1}{2}\right)}^2=0 \end{gathered}$$

Then the roots are which are $m_{1,2}=\frac{1}{2}$real and repeated roots.

Then the solution of the corresponding homogeneous differential equation $y\text{'}\text{'}-y\text{'}+\frac{1}{4}y=0$can be given by

$$\begin{gathered} y_h=c_1{e}^{m_{1}x}+c_{2}x{e}^{m_{2}x} \\ =c_1{e}^{\frac{1}{2}x}+c_{2}x{e}^{\frac{1}{2}x} \end{gathered}$$

Finding the general solution for the non-homogeneous differential equation. 

Secondly, we need to find the particular solution of the non-homogeneous differential equation$y\text{'}\text{'}-y\text{'}+\frac{1}{4}y=3+e^{\frac{x}{2}}$

Assume that$y_p=A+B{x}^2{e}^{\frac{1}{2}x}$is the solution for the non-homogeneous differential equation where $g\left(x\right)=3+e^{\frac{1}{2}x}$, since we have the solution$e^{\frac{1}{2}x},x{e}^{\frac{1}{2}x}$ in the homogeneous solution,

Differentiate the assumption with respect to x, we can obtain,

$y_p\text{'}=2Bx{e}^{\frac{1}{2}x}+\frac{1}{2}B{x}^2{e}^{\frac{1}{2}x}$

Again differentiate another time with respect to x, we can obtain,

$$\begin{gathered} y_p\text{'}\text{'}=2B{e}^{\frac{1}{2}x}+Bx{e}^{\frac{1}{2}x}+Bx{e}^{\frac{1}{2}x}+\frac{1}{4}B{x}^2{e}^{\frac{1}{2}x} \\ =2B{e}^{\frac{1}{2}x}+2Bx{e}^{\frac{1}{2}x}+\frac{1}{4}B{x}^2{e}^{\frac{1}{2}x} \end{gathered}$$

Substituteinto the differential equation, then we can obtain,

$$\begin{gathered} \left[2B{e}^{\frac{1}{2}x}+2Bx{e}^{\frac{1}{2}x}+\frac{1}{4}B{x}^2{e}^{\frac{1}{2}x}\right]-\left[2Bx{e}^{\frac{1}{2}x}+\frac{1}{2}B{x}^2{e}^{\frac{1}{2}x}\right]+\frac{1}{4}\left[A+B{x}^2{e}^{\frac{1}{2}x}\right]=3+e^{\frac{1}{2}x} \\ \frac{1}{4}A+2B{e}^{\frac{1}{2}x}=3+e^{\frac{1}{2}x} \end{gathered}$$

Then by comparison between the left and right side, we can have

$$\begin{gathered} \frac{1}{4}A=3\Rightarrow A=12 \\ 2B=1\Rightarrow B=\frac{1}{2} \end{gathered}$$

Then the assumed solution becomes .$y_p=12+\frac{1}{2}{x}^2{e}^{\frac{1}{2}x}$

From the equations, we can obtain,

$y=c_1{e}^{\frac{1}{2}x}+c_{2}x{e}^{\frac{1}{2}x}+\frac{1}{2}{x}^2{e}^{\frac{1}{2}x}+12$

Thus the general solution of the given differential equation is

$\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\frac{\mathbf{1}}{\mathbf{2}}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{z}}\mathit{x}{\mathit{e}}^{\frac{\mathbf{1}}{\mathbf{2}}\mathbf{x}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{x}}^{\mathbf{2}}{\mathit{e}}^{\frac{\mathbf{1}}{\mathbf{2}}\mathbf{x}}\mathbf{+}\mathbf{12}$