Non-Homogeneous Linear Differential Equation (The Method of Undetermined Coefficients). Solve the equation (D+1)y=sin⁡x

${\displaystyle (D^2+1)y=\sin x}$
${\displaystyle m^2+1=0}$
${\displaystyle m^2=-1}$
${\displaystyle m=\pm i}$
${\displaystyle C.f=C_1\cos x+C_2\sin x}$
${\displaystyle y_t=x(a\cos x+b\sin x)}$ (1)
${\displaystyle y_t{y}_t=\sin x}$ (2)
equation (1) D.w.r. to x
${\displaystyle y_t^{\prime }=x(-a\sin x+b\cos x)+(a\cos x+b\sin x)}$
again D.w.r. to x
${\displaystyle y_{t}x(-a\cos x-b\sin x)+(-a\sin x+v\cos x)\pm a\sin x+b\cos x}$
${\displaystyle =x(-a\cos x-b\sin x)-2a\sin x+2b\cos x}$ (3)
eq (1) and eq (3) put in eq (2)
${\displaystyle x(-a\cos x-b\sin x)-2a\sin x+2b\cos x+x(\cos x+b\sin x)=\sin x}$
equat the coefficient
${\displaystyle -2a\sin x+2b\cos x=\sin x}$
${\displaystyle -2a=1}$
${\displaystyle a=-\frac{1}{2}}$
${\displaystyle 2b=0}$
${\displaystyle b=0}$
P.I is ${\displaystyle y_t=\frac{1}{2}x\cos x}$
${\displaystyle y=c_1\cos x+c_2\sin x-\frac{1}{2}x\sin x}$

Here A.E. is
${\displaystyle m^2+1=0}$ and its roots are ${\displaystyle m=\pm i}$
Hence ${\displaystyle C.F.=C_1\cos x+C_2\sin x}$
Note that sin x is common in the C.F. and the R.H.S. of the given equation. (${\displaystyle \pm }$ i is the root of the A.E.)
Therefore P.I. is y the form ${\displaystyle yp=x(a\cos x+b\sin x)}$ ...(1)
Since ${\displaystyle \pm }$ i is root of the A.E.
We have to find a and b such that ${\displaystyle y_p+y_p=\sin x}$ ...(2)
From Eqn. (1) ${\displaystyle y_p^{\prime }=x(-a\sin x+b\cos x)+(a\cos x+b\sin x)}$
${\displaystyle y_p=x(-a\cos x-b\sin x)+(-a\sin x+b\cos x)-a\sin x+b\cos x}$
${\displaystyle =x(-a\cos x-b\sin x)-2a\sin x+2b\cos x}$
Then the given equation reduces to using the Eqn. (1) ${\displaystyle x(-a\cos x-b\sin x)-2a\sin x+2b\cos x+x(a\cos x+b\sin x)=\sin x}$
Equating the coefficients, we get
${\displaystyle i.e.,-2a\sin x+2b\cos x=\sin x}$
${\displaystyle -2a=1,2b=0}$
${\displaystyle a=\frac{-1}{2},b=0}$
Thus P.I. is ${\displaystyle y_p=\frac{-1}{2}x\cos x}$
${\displaystyle y=C.F.+P.I.}$
${\displaystyle =C_1\cos x+C_2\sin x-\frac{1}{2}x\sin x}$.

$\begin{array}{}Given:(D^2+1)y=\sin x\\ P.I.=\frac{1}{D^2+1}\sin x\\ =\frac{1}{-1+1}\sin x\\ =x\frac{1}{2D}\sin x\\ =\frac{x}{2}\int \sin xdx\\ =\frac{x}{2}(-\cos x)\\ P.I.=\frac{-x\cos x}{2}\end{array}$