Question

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

$\mathit{y}\mathbf{+}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathit{y}\mathbf{=}\mathbf{2}\mathit{x}$

Short answer

$y=c_1{e}^{2x}+c_2{e}^{-3x}-\frac{1}{3}x-\frac{1}{18}$

Step-by-step solution

General solution for homogeneous differential equation

We have the differential equation

$y+y\text{'}-6y=2x$

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.

$\begin{array}{c}{m}^2{e}^{mx}+m{e}^{mx}-6e^{mx}=0\\ e^{mx}\left(m^2+m-6\right)=0\end{array}$

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

$\begin{array}{c}{m}^2+m-6=0\\ \left(m-2\right)\left(m+3\right)=0\end{array}$

Then the roots are

$m_1=2 , m_2=-3$

which are real and distinct roots.

Then the solution of the corresponding homogeneous equation $y+y\text{'}-6y=0$ can be given by $y_h=c_1{e}^{2x}+c_2{e}^{-3x}$

General solution for non-homogeneous differential equation

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

equation$y+y\text{'}-6y=2x$as the following technique :

Assume that$y_p=Ax+B$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 $y_p=Ax+B$ into the non-homogeneous differential equation, then we obtain

$\begin{array}{c}0+A-6Ax-6B=2x\\ -6Ax+\left(A-6B\right)=2x\end{array}$

Then we have

$\begin{array}{l}A=-\frac{1}{3}\\ B=-\frac{1}{18}\end{array}$

Then the particular solution becomes

$y_p=-\frac{1}{3}x-\frac{1}{18}$

Then we obtain

$\begin{array}{c}y=y_h+y_p\\ =c_1{e}^{2x}+c_2{e}^{-3x}-\frac{1}{3}x-\frac{1}{18}\end{array}$

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$y=c_1{e}^{2x}+c_2{e}^{-3x}-\frac{1}{3}x-\frac{1}{18}$.