91Ó°ÊÓ

Given equation is$(D^2+1)y=8x\sin x$

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 differential equation is

$(D^2+1)y=8x\sin x$

The auxiliary equation can be written as$⇒m^2+1=0$

The roots of the equation are

data-custom-editor="chemistry" $⇒m=Â\pm i$

The complementary function is

$\begin{array}{c}\text{C.}F=C_1\sin x+C_2\cos x\text{P.I}\\ y=x{e}^{ix}(Ax+B)\\ y^{\text{'}}=e^{ix}(B+iBx+A(2+ix\left)x\right)\\ y^{\text{'}\text{'}}=−e^{ix}(B(x−2i)+A(x^2−4ix−2)\end{array}$

Hence,

$\begin{array}{l}−e^{ix}(B(x−2i)+A(x^2−4ix−2)=8xi{e}^{ix}\\ \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}A+iB+2iAx=8xi{e}^{ix}\end{array}$

Solve the equation further,

$\begin{array}{c}A=−2i,B=2\\ y_p=2x{e}^{ix}(1−ix)\\ y_p=2x(\sin x−x\cos x)\end{array}$

Hence the equation is

$\begin{array}{l}\text{P.I}=2x(\sin x−x\cos x)\\ \text{C.}S=C_1\sin x+C_2\cos x+2x(\sin x−x\cos x)\end{array}$

The solution of the differential equation is

$y\left(x\right)=C_1\sin x+C_2\cos x+2x(\sin x−x\cos x)$