Question

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

$\frac{\mathbf{1}}{\mathbf{4}}\mathit{y}\mathbf{+}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathit{x}$

Short answer

$y=c_1{e}^{-2x}+c_{2}x{e}^{-2x}+x^2-4x+\frac{7}{2}$

Step-by-step solution

General solution for homogeneous differential equation

We have the differential equation

$\frac{1}{4}y+y\text{'}+y=x^2-2x$

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 to obtain the auxiliary equation.

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

Since $e^{mx}$cannot be equal to 0 , then we have

$\begin{array}{c}\frac{1}{4}{m}^2+m+1=0\\ m^2+4m+4=0\\ {\left(m+2\right)}^2=0\end{array}$

Then the roots are

$m_{1,2}=-2$

which are real and repeated.

Since we have repeated roots$m_{1,2}=-2$, then we need to have two linearly independent solutions, then we obtain the homogeneous solution as

$\begin{array}{c}{y}_h=n{e}^{-2x}\left(k+lx\right)\\ =c_1{e}^{-2x}+c_{2}x{e}^{-2x}\end{array}$

where we have $c_1=nk$and $c_2=nl$.

General solution for non-homogeneous differential equation

Second, we have to find the particular solution of the non-homogeneous differential

equation$\frac{1}{4}y+y\text{'}+y=x^2-2x$as the following technique :

Assume that$y_p=A{x}^2+Bx+D$is a solution for the non-homogeneous differential

equation.

After that, differentiate the assumption with respect to x, then we obtain

$y_p\text{'}=A,y_p\text{'}\text{'}=0$

Substitute with $y_p=A{x}^2+Bx+D$ into the non-homogeneous differential equation, then we obtain

$\begin{array}{c}\frac{1}{2}A+\left(2Ax+B\right)+\left(A{x}^2+Bx+D\right)=x^2-2x\\ A{x}^2+\left(2A+B\right)x+\left(\frac{1}{2}A+B+D\right)=x^2-2x\end{array}$

Then we have

$\begin{array}{c}A=1\\ 2A+B=-2 \Rightarrow B=-4\\ \frac{1}{2}A+B+D=0 \Rightarrow D=\frac{7}{2}\end{array}$

Then the particular solution becomes

$y_p=x^2-4x+\frac{7}{2}$

Then we obtain

$\begin{array}{c}y=y_h+y_p\\ =c_1{e}^{-2x}+c_{2}x{e}^{-2x}+x^2-4x+\frac{7}{2}\end{array}$

is the general solution of the non-homogeneous differential equation.

Therefore,the generalsolution of the given differential equation by undetermined coefficients is found to be$y=c_1{e}^{-2x}+c_{2}x{e}^{-2x}+x^2-4x+\frac{7}{2}$.