Question

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

$y\text{'}\text{'}+3y\text{'}+2y=6$

Short answer

$y=c_1 e^{-x} + c_2 e^{-2x}+3$

Step-by-step solution

Solving equation

We have the non-homogeneous differential equation

$y\text{'}\text{'}+3y\text{'}+2y=6$

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$y\text{'}\text{'}+3y\text{'}+2y=0$to obtain the auxiliary equation.

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

For any real,$x,e^{mx}\ne 0$ we have

$\begin{array}{c} m^2+3m+2=0\\ \left(m+1\right) \left(m+2\right)=0\end{array}$

Then, the roots as $m_1=-1\mathrm{and}{m}_2=-2$ which are real and distinct.

Then the solution of the corresponding homogeneous equation $y\text{'}\text{'}+3y\text{'}+2y=0$can be given asrole="math" localid="1663846868410" ${\mathit{y}}_{\mathbf{h}}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathbf{ }{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{ }{\mathit{c}}_{\mathbf{2}}\mathbf{ }{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }$

Solution of the non-homogeneous differential equation

$y\text{'}\text{'}+3y\text{'}+2y=6$Second, we have to find the particular solution of the non-homogeneous differential equationas the following technique:

Assume that $y_p=A{e}^x+B$is a solution for the non-homogeneous differential equation.

After that, differentiate the assumption with respect to x, and substitute $y{\text{'}}_p=A{e}^x,y\text{'}{\text{'}}_p=A{e}^x$into $y\text{'}\text{'}+3y\text{'}+2y=6$to obtain the auxiliary equation.

$\begin{array}{c}A{e}^x+3A{e}^x+2A{e}^x+2B=6 \\ 6A{e}^x+2B=6\\ A=0 B=3\end{array}$

Then the particular solution becomes$y_p=3$

Then we obtain

role="math" localid="1663846781548" $\begin{array}{l}y=y_h+y_p\\ =c_1 e^{-x} + c_2 e^{-2x}+3\end{array}$

The general solution of the non-homogeneous differential equation

$y=c_1 e^{-x} + c_2 e^{-2x}+3$

Hence the final solution is $y=c_1 e^{-x} + c_2 e^{-2x}+3$