Hello Muhammad Fasih,
We are given the following ODE to solve:
$$y''+4y=xe^x+x\sin(2x)$$
Rather than referring to a table to determine the form of the particular solution, let's employ the annihilator method instead.
Let's look at the first term on the right side:
$$f(x)=xe^x$$
Differentiating, we find:
$$f'(x)=xe^x+e^x$$
$$f''(x)=xe^x+2e^x$$
Now, observing:
$$f''(x)-2f'(x)+f(x)=xe^x+2e^x-2\left(xe^x+e^x \right)+xe^x=0$$
We may conclude the differential operator:
$$A\equiv(D-1)^2$$
annihilates $f$.
Now, let's look at the second term on the right of the ODE.
$$g(x)=x\sin(2x)$$
Differentiating, we find:
$$g''(x)=4\left(\cos(2x)-x\sin(2x) \right)$$
$$g^{(4)}(x)=16\left(x\sin(2x)-2\cos(2x) \right)$$
Now, observing:
$$g^{(4)}(x)+8g''(x)+16g(x)=16\left(x\sin(2x)-2\cos(2x) \right)+32\left(\cos(2x)-x\sin(2x) \right)+16x\sin(2x)=0$$
We may conclude the differential operator:
$$A\equiv\left(D^2+4 \right)^2$$
annihilates $g$.
Hence, the operator:
$$C\equiv(D-1)^2\left(D^2+4 \right)^2$$
annhilates $f(x)+g(x)$.
Applying the operator to the ODE, we obtain:
$$\left((D-1)^2\left(D^2+4 \right)^3 \right)[y]=0$$
And for this homogeneous ODE with repeated characteristic roots, we obtain the general solution:
$$y(x)=\left(c_1+c_2x \right)e^x+\left(c_3+c_4x+c_5x^2 \right)\cos(2x)+\left(c_6+c_7x+c_8x^2 \right)\sin(2x)$$
Now, we know the solution to the original ODE will be the superposition of the homogenous and particular solutions. If we note that the homogenous solution is of the form:
$$y_h(x)=c_3\cos(2x)+c_6\sin(2x)$$
We may therefore conclude that the particular solution must have the form:
$$y_p(x)=\left(c_1+c_2x \right)e^x+\left(c_4x+c_5x^2 \right)\cos(2x)+\left(c_7x+c_8x^2 \right)\sin(2x)$$
Differentiating twice, and substituting into the original ODE, we obtain:
$$\left(5c_1+2c_2+5c_2x \right)e^x+\left(2c_5+4c_7+8c_8x \right)\cos(2x)+\left(-4c_4+2c_8-8c_5x \right)\sin(2x)=\left(0+1x \right)e^x+\left(0+0x \right)\cos(2x)+\left(0+1x \right)\sin(2x)$$
Equating coefficients, we obtain the system:
$$5c_1+2c_2=0$$
$$5c_2=1$$
$$2c_5+4c_7=0$$
$$8c_8=0$$
$$-4c_4+2c_8=0$$
$$-8c_5=1$$
From this, we obtain:
$$\left(c_1,c_2,c_4,c_5,c_7,c_8 \right)=\left(-\frac{2}{25},\frac{1}{5},0,-\frac{1}{8},\frac{1}{16},0 \right)$$
And so our particular solution is:
$$y_p(x)=\left(-\frac{2}{25}+\frac{1}{5}x \right)e^x+\left(-\frac{1}{8}x^2 \right)\cos(2x)+\left(\frac{1}{16}x \right)\sin(2x)$$
$$y_p(x)=\frac{5x-2}{25}e^x+\frac{x}{16}\left(\sin(2x)-2x\cos(2x) \right)$$
And so, the general solution to the given ODE is:
$$y(x)=y_h(x)+y_p(x)=c_1\cos(2x)+c_2\sin(2x)+\frac{5x-2}{25}e^x+\frac{x}{16}\left(\sin(2x)-2x\cos(2x) \right)$$