Find the general solution of the following differential equations (complementary function + particular solution). Find the particular solution by inspection or by $(\mathbf{6}.\mathbf{18}),(\mathbf{6}.\mathbf{23}),$or.$\mathbf{(}\mathbf{6}\mathbf{.24}\mathbf{)}$ Alsofind a computer solution and reconcile differences if necessary, noticing especially whether the particular solution is in simplest form [see$\mathbf{(}\mathbf{6}\mathbf{.26}\mathbf{)}$and the discussion after$\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{15}\mathbf{)}$].

Hint: First solve.

Given equation is$(D^2+2D+17)y=60e^{-4x}\sin 5x$

${(D-2)}^{2}y=16$

A general solution to the nth order differential equation is one that incorporates a significant number of arbitrary constants. If one uses the variable approach to solve a first-order differential equation, one must insert an arbitrary constant as soon as integration is completed.

The given equation is

$(D^2+2D+17)y=60e^{-4x}\sin 5x$

The auxiliary equation can be written as

$\begin{array}{l}\Rightarrow m^2+2m+17=0\\ \Rightarrow m=\frac{-2\pm \sqrt{4-68}}{2}\\ \Rightarrow m=-1\pm 4i\end{array}$

The complementary function can be written as below

$C.F=(C_1\sin 4x+C_2\cos 4x)e^{-x}$

Putting Das $\text{(D-4)}$as power of the exponential here is-4

$P.I=\frac{1}{D^2+2D+17}60{\text{e}}^{-4x}\sin 5x$

Solve the equation further

$\begin{array}{l}\Rightarrow \frac{1}{{(D-4)}^2+2(D-4)+17}60{\text{e}}^{-4x}\sin 5x\\ \Rightarrow \frac{1}{D^2-8D+16+2D-8+17}60e^{-4x}\sin 5x\\ \Rightarrow \frac{1}{D^2-6D+25}60{\text{e}}^{-4x}\sin 5x\\ \Rightarrow \frac{1}{-25-6D+25}60{\text{e}}^{-4x}\sin 5x\end{array}$

Putting $D^2=-a^2$denominator becomes0

$\begin{array}{l}\Rightarrow -\frac{1}{6D}60{\text{e}}^{-4x}\sin 5x(\frac{1}{D}=\text{integration})\Rightarrow -10\int \sin 5x{e}^{-4x}dx\\ \Rightarrow -10\{\sin 5x\int e^{-4x}dx-\int (5\cos 5x\int e^{-4x}dx)dx\}\\ \Rightarrow -10\{-\sin 5x\frac{e^{-4x}}{4}+\int 5\cos 5x\frac{e^{-4x}}{4}dx\\ \Rightarrow \frac{-160}{41}\{-\frac{1}{4}\sin 5x{e}^{-4x}+\frac{5}{16}\cos 5x{e}^{-4x}\end{array}$

Now the answer is,

$P.I=\frac{-160}{41}\{-\frac{1}{4}\sin 5x{e}^{-4x}+\frac{5}{16}\cos 5x{e}^{-4x}\}$

Hence,

$\begin{array}{l}C\cdot S=(C_1\sin 4x+C_2\cos 4x)e^{-x}+\frac{-160}{41}\{-\frac{1}{4}\sin 5x{e}^{-4x}+\frac{5}{16}\cos 5x{e}^{-4x}\}\\ y\left(x\right)=(C_1\sin 4x+C_2\cos 4x)e^{-x}+\frac{-160}{41}\{-\frac{1}{4}\sin 5x{e}^{-4x}+\frac{5}{16}\cos 5x{e}^{-4x}\}\end{array}$