Question: Identify each of the differential equations in Problems 1to 24 as to type (for example, separable, linear first order, linear second order, etc.), and then solve it.

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}\mathbf{5}\mathit{x}\mathbf{+}\mathbf{4}{\mathit{e}}^{\mathbf{x}}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{s}\mathit{i}\mathit{n}\mathbf{2}\mathit{x}\mathbf{)}$

The given function is$y^{\text{'}\text{'}}-2y^{\text{'}}+5y=5x+4e^x(1+\sin 2x)$

When fand its derivatives are inserted into the equation, a solution is a function y = f (x) that solves the differential equation. The highest order of any derivative of the unknown function appearing in the equation is the order of a differential equation.

A differential equation of the form$\mathbf{(}\mathit{D}\mathbf{-}\mathit{a}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{-}\mathit{b}\mathbf{)}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{a}\mathbf{\ne }\mathit{b}$ has general solution$\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{a}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{b}\mathbf{x}}\mathbf{.}$

Let the solution is.

$y=y_c+y_p$

Where, y_c is complementary function and y_p is known as particular solution.

$\begin{array}{c}{y}^{\text{'}\text{'}}-2y^{\text{'}}+5y=5x+4e^x(1+\sin 2x)\\ \left(D^2-2D+5\right)y=5x+4e^x(1+\sin 2x)\\ \left(D^2-2D+5\right)y=5x+4e^x+4e^x(1+\sin 2x)\end{array}$

The auxiliary equation of above equation is.

$m^2-2m+5=0$

On comparing with the quadratic equation $a{m}^2+bm+c$.

$\begin{array}{c}a=1\\ b=-2\\ c=5\end{array}$

The root of quadratic equation is calculated by using formula,

$m=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

The roots of equation are.

$\begin{array}{l}m=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ m=\frac{-(-2)\pm \sqrt{{(-2)}^2-4\left(1\right)\left(5\right)}}{2\left(1\right)}\\ m=\frac{2\pm \sqrt{4-20}}{2}\\ m=\frac{2\pm 4i}{2}\end{array}$

$m=1\pm 2i$

It can be written as,

$m=1+2i\text{and}m=1-2i$

The real part is represented by a and complex part by b.

$\begin{array}{l}a+ib=1+2i\\ a-ib=1-2i\end{array}$

The complementary function y_c of differential equation with complex root is.

$y_c=e^{ax}\left(A{e}^{2ik}+B{e}^{-2ik}\right)$

A and B are arbitrary constants.

$$\begin{gathered} \begin{array}{l}{e}^{aix}=\cos ax+i\sin ax\\ y_c=e^x\left(A{e}^{2ix}+B^{-2ik}\right)\\ y_c=e^x\left(A\right(\cos 2x+i\sin 2x)+B(\cos (-2)x+i\sin (-2)x\left)\right)\\ y_c=e^x\left(\right(A+B)\cos 2x+[i(A-B)\left]\sin 2x\right)\end{array} \\ {y}_c=e^x\left(c_1\cos 2x+c_2\sin 2x\right) \end{gathered}$$

Therefore,

$\begin{array}{l}{c}_1=(A+B)\\ c_1=i(A-B)\end{array}$

We have,

$\begin{array}{c}=5x+4e^x+4e^x\sin 2x\\ Q=Q_1+Q_2+Q_3\end{array}$

The complete particular solution in this case is.

$\begin{array}{l}{y}_p=y_{p1}+y_{p2}\\ Q_1=5x\end{array}$

y_p1 Is calculated as.

$\begin{array}{l}{y}_{p1}=\frac{1}{P\left(D^2-2D+5\right)}\left(5x\right)\\ y_{p1}=\frac{1}{5\left(\frac{D^2}{5}-\frac{2D}{5}+1\right)}\left(5x\right)\\ y_{p1}=\left(1-\left(\frac{D^2}{5}-\frac{2D}{5}\right)+{\left(\frac{D^2}{5}-\frac{2D}{5}\right)}^2+--\right)\left(x\right)\\ y_{p1}=x+\frac{2}{5}\end{array}$

Also,

$Q_2=4e^x$

The formula for calculating the particular solution is.

$\begin{array}{c}{y}_p=\frac{b{e}^{ax}}{P\left(a\right)}\\ P\left(a\right)\ne 0\end{array}$

The particular solution is calculating as,

$\begin{array}{l}{y}_{p1}=\frac{1}{P\left(D\right)}4e^x\\ y_{p1}=\frac{1}{D^2-2D+5}{e}^x\\ y_{p1}=\frac{4}{4}{e}^x\\ y_{p1}=e^x\end{array}$

Also,

$\begin{array}{l}{Q}_3=4e^x\sin 2x\\ e^{aix}=\cos ax+i\sin ax\end{array}$

This implies.

$\begin{array}{l}\mathrm{Re}\left(e^{aix}\right)=\cos ax\\ \mathrm{Re}\left(e^{aix}\right)=i\sin ax\end{array}$

We can rewrite,

$\begin{array}{l}{Q}_3=4e^x\sin 2x\\ Q_3=4e^{(1+2i)x}\end{array}$

The formula for calculating the particular solution is.

$\begin{array}{c}{y}_p=\frac{b{e}^{iax}}{P(c+ia)}\\ P(c+ia)\ne 0\end{array}$

Now,

$\begin{array}{l}P(1+2i)=\left({(1+2i)}^2-2(1+2i)+5\right)\\ P(1+2i)=0\end{array}$

The particular solution is calculating as.

$\begin{array}{l}{y}_{p1}=\frac{1}{P\left(D\right)}4e^{(1+2i)x}\\ y_{p1}=\frac{1}{D^2+2D+5}{e}^{(1+2i)x}\\ y_{p1}=\frac{4e^{(1+2i)x}}{4iD\left(\frac{D}{4i}+1\right)}\left(1\right)\\ y_{p1}=\frac{-i{e}^{(1+2i)x}}{D}\left(1-\frac{D}{4i}+{\left(\frac{D}{4i}\right)}^2---\right)\end{array}$

The complete solution for y_p is,

$y_p=x-\frac{2}{5}+e^x-x{e}^x\cos 2x$

Therefore, the complete solution is,

$\begin{array}{l}y=y_c+y_p\\ y=e^x\left(A{e}^{2ix}+B{e}^{-2ix}\right)+x+\frac{2}{5}+e^x-x{e}^x\cos 2x\end{array}$

Thus, the general solution of equation $y^{\text{'}\text{'}}-2y^{\text{'}}+5y=5x+4e^x(1+\sin 2x)$is

$y=e^x\left(A{e}^{2ix}+B{e}^{-2ix}\right)+x+\frac{2}{5}+e^x-x{e}^x\cos 2x.$