91Ó°ÊÓ

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

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

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

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

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

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

Now the answer is,

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

Hence,

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