Find a particular solution to the differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)\mathbf{+}\mathbf{y}\left(\mathbf{x}\right)\mathbf{=}\mathbf{4}\mathbf{xcosx}$

The given differential equation is;

$\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{y}\left(\mathrm{x}\right)=4\mathrm{xcosx}\dots \left(1\right)$

Write the homogeneous differential equation of the equation (1),

$\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{y}\left(\mathrm{x}\right)=0$

The auxiliary equation for the above equation,

${\mathrm{m}}^2+1=0$

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^2+1=0\\ \mathrm{m}=\pm \mathrm{i}\end{array}$

The roots of the auxiliary equation are,

${\mathrm{m}}_1=\mathrm{i},\&{\mathrm{m}}_2=-\mathrm{i}$

The complementary solution of the given equation is:

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1\mathrm{cosx}+{\mathrm{c}}_2\mathrm{sinx}$

According to the method of undetermined coefficients, assume, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}={\mathrm{Ax}}^2\mathrm{cosx}+\mathrm{Bxcosx}+{\mathrm{Cx}}^2\mathrm{sinx}+\mathrm{Dxsinx}\\ {\mathrm{y}}_{\mathrm{p}}=({\mathrm{Ax}}^2+\mathrm{Bx})\mathrm{cosx}+({\mathrm{Cx}}^2+\mathrm{Dx})\mathrm{sinx}......\left(2\right)\end{array}$

Now find the derivative of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}=-({\mathrm{Ax}}^2+\mathrm{Bx})\mathrm{sinx}+(2\mathrm{Ax}+\mathrm{B})\mathrm{cosx}+({\mathrm{Cx}}^2+\mathrm{Dx})\mathrm{cosx}+(2\mathrm{Cx}+\mathrm{D})\mathrm{sinx}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}=(-{\mathrm{Ax}}^2-\mathrm{Bx}+2\mathrm{Cx}+\mathrm{D})\mathrm{sinx}+(2\mathrm{Ax}+\mathrm{B}+{\mathrm{Cx}}^2+\mathrm{Dx})\mathrm{cosx}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}=(-{\mathrm{Ax}}^2-\mathrm{Bx}+2\mathrm{Cx}+\mathrm{D})\mathrm{cosx}+(2\mathrm{Ax}+\mathrm{B}+{\mathrm{Cx}}^2+\mathrm{Dx})(-\mathrm{sinx})\\ +(-2\mathrm{Ax}-\mathrm{B}+2\mathrm{C})\mathrm{sinx}+(2\mathrm{A}+2\mathrm{Cx}+\mathrm{D})\mathrm{cosx}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}=(-{\mathrm{Ax}}^2-\mathrm{Bx}+2\mathrm{Cx}+\mathrm{D}+2\mathrm{A}+2\mathrm{Cx}+\mathrm{D})\mathrm{cosx}+(-2\mathrm{Ax}-\mathrm{B}-{\mathrm{Cx}}^2-\mathrm{Dx}-2\mathrm{Ax}-\mathrm{B}+2\mathrm{C})\mathrm{sinx}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}=(-{\mathrm{Ax}}^2-\mathrm{Bx}+4\mathrm{Cx}+2\mathrm{D}+2\mathrm{A})\mathrm{cosx}+(-4\mathrm{Ax}-2\mathrm{B}-{\mathrm{Cx}}^2-\mathrm{Dx}+2\mathrm{C})\mathrm{sinx}\end{array}$

From the equation (1), Substitute the value of ${\mathbf{y}}_p\mathbf{\text{'}}\mathbf{\text{'}}$and${\mathbf{y}}_p$ in the equation (1),

$\begin{array}{l}\Rightarrow {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{x}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=4\mathrm{xcosx}\\ \Rightarrow (-{\mathrm{Ax}}^2-\mathrm{Bx}+4\mathrm{Cx}+2\mathrm{D}+2\mathrm{A})\mathrm{cosx}+(-4\mathrm{Ax}-2\mathrm{B}-{\mathrm{Cx}}^2-\mathrm{Dx}+2\mathrm{C})\mathrm{sinx}\\ +({\mathrm{Ax}}^2+\mathrm{Bx})\mathrm{cosx}+({\mathrm{Cx}}^2+\mathrm{Dx})\mathrm{sinx}=4\mathrm{xcosx}\\ \Rightarrow (4\mathrm{Cx}+2\mathrm{D}+2\mathrm{A})\mathrm{cosx}+(-4\mathrm{Ax}-2\mathrm{B}+2\mathrm{C})\mathrm{sinx}=4\mathrm{xcosx}\\ \Rightarrow \left(4\mathrm{Cx}\right)\mathrm{cosx}+(2\mathrm{D}+2\mathrm{A})\mathrm{cosx}-\left(4\mathrm{Ax}\right)\mathrm{sinx}+(-2\mathrm{B}+2\mathrm{C})\mathrm{sinx}=4\mathrm{xcosx}\end{array}$

Comparing all coefficients of the above equation;

$\begin{array}{c}4\mathrm{C}=4\Rightarrow \mathrm{C}=1\\ -4\mathrm{A}=0\Rightarrow \mathrm{A}=0\\ 2\mathrm{D}+2\mathrm{A}=0.......\left(3\right)\\ 2\mathrm{C}-2\mathrm{B}=0........\left(4\right)\end{array}$

Substitute the value of C in the equation (4),

$\begin{array}{c}2\left(1\right)-2\mathrm{B}=0\\ \mathrm{B}=1\end{array}$

Substitute the value of A in the equation (3),

$\begin{array}{c}2\mathrm{D}+2\left(0\right)=0\\ \mathrm{D}=0\end{array}$

Substitute the value of A, B, Cand Din the equation (2),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}=({\mathrm{Ax}}^2+\mathrm{Bx})\mathrm{cosx}+({\mathrm{Cx}}^2+\mathrm{Dx})\mathrm{sinx}\\ {\mathrm{y}}_{\mathrm{p}}=(\left(0\right){\mathrm{x}}^2+\left(1\right)\mathrm{x})\mathrm{cosx}+(\left(1\right){\mathrm{x}}^2+\left(0\right)\mathrm{x})\mathrm{sinx}\\ {\mathrm{y}}_{\mathrm{p}}=\mathrm{xcosx}+{\mathrm{x}}^2\mathrm{sinx}\end{array}$

Therefore, the particular solution of equation (1),

${\mathbf{y}}_p\mathbf{=}\mathbf{xcosx}\mathbf{+}{\mathbf{x}}^2\mathbf{sinx}$