Question

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

$\mathit{y}\mathbf{-}\mathbf{8}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{20}\mathit{y}\mathbf{-}\mathbf{100}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{26}\mathit{x}{\mathit{e}}^{\mathbf{x}}$

Short answer

$y=k_1{e}^{4x}\cos 2x+k_2{e}^{4x}\sin 2x+5x^2+4x+\frac{11}{10}-2x{e}^x-\frac{12}{13}{e}^x$

Step-by-step solution

General solution for homogeneous differential equation

We have the differential equation

$y-8y\text{'}+20y=100x^2-26x{e}^x$

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 $y-8y\text{'}+20y=0$ to obtain the auxiliary equation.

$\begin{array}{l}{m}^2{e}^{mx}-8m{e}^{mx}+20e^{mx}=0\\ e^{mx}\left(m^2-8m+20\right)=0\end{array}$

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

$m^2-8m+20=0$

Then the roots are

$m_{1,2}=\frac{8\pm \sqrt{64-4\times 20}}{2}=4\pm 2i$

which are complex roots..

Then the solution of the corresponding homogeneous equation $y-8y\text{'}+20y=0$can be given by

$\begin{array}{c}{y}_h=c_1{e}^{\left(4+2i\right)x}+c_2{e}^{\left(4-2i\right)x}\\ =c_1{e}^{4x}\left(\cos 2x+\sin 2x\right)+c_2{e}^{4x}\left(\cos 2x-\sin 2x\right)\\ =\left(c_1+c_2\right)e^{4x}\cos 2x+\left(c_1-c_2\right)e^{4x}\sin 2x\\ =k_1{e}^{4x}\cos 2x+k_2{e}^{4x}\sin 2x\end{array}$

General solution for non-homogeneous differential equation

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

equation $y-8y\text{'}+20y=100x^2-26x{e}^x$as the following technique :

Assume that $y_p=A{x}^2+Bx+C+Dx{e}^x+E{e}^x$ is a solution for the non-homogeneous differential equation where $g\left(x\right)=100x^2-26x{e}^x$.

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

$\begin{array}{c}{y}_p\text{'}=2Ax+B+D{e}^x+Dx{e}^x+E{e}^x\\ =2Ax+B+D\left(x+1\right)e^x+E{e}^x\end{array}$

Differentiate another time with respect to x, then we obtain

$\begin{array}{c}{y}_p\text{'}\text{'}=2A+D\left(x+1\right)e^x+D{e}^x+E{e}^x\\ y_p\text{'}\text{'}=2A+D\left(x+2\right)e^x+E{e}^x\end{array}$

Substitute $y-8y\text{'}+20y=100x^2-26x{e}^x$ into the non-homogeneous differential equation $y_p=A{x}^2+Bx+C+Dx{e}^x+E{e}^x$, then we obtain

$\begin{array}{l}\left[2A+D\left(x+2\right)e^x+E{e}^x\right]-8\left[2Ax+B+D\left(x+1\right)e^x+E{e}^x\right]+\\ 20\left[A{x}^2+Bx+C+Dx{e}^x+E{e}^x\right]=100x^2-26x{e}^x\\ \Rightarrow 20A{x}^2+\left(20B-16A\right)x+\left(2A-8B+20C\right)+\\ 13Dx{e}^x+\left(13E-6D\right)e^x=100x^2-26x{e}^x\end{array}$

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

$\begin{array}{l}20A=100 \Rightarrow A=1\\ 20B-16A=0 \Rightarrow B=4\\ 2A-8B+20C=0 \Rightarrow C=\frac{11}{10}\\ 13D=-26 \Rightarrow D=-2\\ 13E-6D=0 \Rightarrow E=-\frac{12}{13}\end{array}$

Then the particular solution becomes

$y_p=5x^2+4x+\frac{11}{10}-2x{e}^x-\frac{12}{13}{e}^x$

Then we obtain

$\begin{array}{l}y=y_h+y_p\\ =k_1{e}^{4x}\cos 2x+k_2{e}^{4x}\sin 2x+5x^2+4x+\frac{11}{10}-2x{e}^x-\frac{12}{13}{e}^x\end{array}$

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

Therefore, the general solution of the given differential equation by undetermined coefficients is found to be

$y=k_1{e}^{4x}\cos 2x+k_2{e}^{4x}\sin 2x+5x^2+4x+\frac{11}{10}-2x{e}^x-\frac{12}{13}{e}^x$