Question

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

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathit{x}\mathbf{+}\mathbf{5}\mathbf{-}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}$

Short answer

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

Step-by-step solution

Finding the corresponding homogeneous equation

$e^{mx}$We are given with the non-homogeneous differential equation$y\text{'}\text{'}+2y\text{'}=2x+5-e^{-2x}$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,$y\text{'}\text{'}+2y\text{'}=0$

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

localid="1663829916152" $\begin{array}{c}{\mathbf{m}}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{m}\mathbf{x}}\mathbf{+}\mathbf{2}\mathbf{m}{\mathbf{e}}^{\mathbf{m}\mathbf{x}}\mathbf{=}\mathbf{0}\\ {\mathbf{e}}^{\mathbf{m}\mathbf{x}}\left(m^2+2m\right)\mathbf{=}\mathbf{0}\end{array}$

Since cannot be equal to 0, then we have

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

Then the roots are $m_1=0,m_2=-2$which are complex roots.

Then the solution of the corresponding homogeneous differential equation $y\text{'}\text{'}+2y\text{'}=0$can be given by

$y_h=c_1{e}^0+c_2{e}^{-2x}$

Finding the general solution for the non-homogeneous differential equation

Secondly, we need to find the particular solution of the non-homogeneous differential equation$y\text{'}\text{'}+2y\text{'}=2x+5-e^{-2x}$

Assume that$y_p=Ax{e}^{-2x}+B{x}^2+Cx$is the solution for the non-homogeneous differential equation where $g\left(x\right)=2x+5-e^{-2x}$, and we timed $e^{-2x}$by x, because $e^{-2x}$is already described in the homogeneous solution above

$c_1+c_2{e}^{-2x}$

Differentiate the assumption with respect tox, we can obtain,

$y_p\text{'}=A{e}^{-2x}-2Ax{e}^{-2x}+2Bx+C$

Again differentiate another time with respect tox, we can obtain,

$\begin{array}{c}{y}_p\text{'}\text{'}=-2A{e}^{-2x}+4Ax{e}^{-2x}-2A{e}^{-2x}+2B\\ =4Ax{e}^{-2x}-4A{e}^{-2x}+2B\end{array}$

Substitute into the differential equation, then we can obtain,

$\begin{array}{c}\left(4Ax{e}^{-2x}-4A{e}^{-2x}+2B\right)+2\left(A{e}^{-2x}-2Ax{e}^{-2x}+2Bx+C\right)=2x+5-e^{-2x}\\ 4Bx+\left(2B+2C\right)-2A{e}^{-2x}=2x+5-e^{-2x}\\ A=\frac{1}{2},B=\frac{1}{2}\\ 2B+2C=5\Rightarrow C=2\end{array}$

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

From the equations, we can obtain,

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

Thus the general solution of the given differential equation is

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