Find a general solution to the differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{sin\theta }\mathbf{-}\mathbf{cos\theta }$

The differential equation is,

$\mathrm{y}\text{'}\text{'}+4\mathrm{y}=\mathrm{sin\theta }-\mathrm{cos\theta }......\left(1\right)$

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

$\mathrm{y}\text{'}\text{'}+4\mathrm{y}=0$

The auxiliary equation for the above equation,

${\mathrm{m}}^2+4=0$

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^2+4=0\\ \mathrm{m}=\pm 2\mathrm{i}\end{array}$

The roots of the auxiliary equation are,

${\mathrm{m}}_1=2\mathrm{i},{\mathrm{m}}_2=-2\mathrm{i}$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1\cos \left(2\mathrm{\theta }\right)+{\mathrm{c}}_2\sin \left(2\mathrm{\theta }\right)$

According to the method of undetermined coefficients, to find a particular solution to the differential equation;

$\mathrm{ay}\text{'}\text{'}+\mathrm{by}\text{'}+\mathrm{cy}=\left\{\begin{array}{l}{\mathrm{Ct}}^{\mathrm{m}}{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}\\ \mathrm{or}\\ {\mathrm{Ct}}^{\mathrm{m}}{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}\end{array}\right\}......\left(2\right)$

For $\mathrm{\beta }\ne 0$, use the form

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{t}}^{\mathrm{s}}[({\mathrm{A}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}+{\mathrm{t}}^{\mathrm{s}}({\mathrm{B}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{B}}_1\mathrm{t}+{\mathrm{B}}_0){\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}]$

With s = 1 if$\mathrm{\alpha }+\mathrm{i\beta }$ is a root of the associated auxiliary equation.

And s = 0 if $\mathrm{\alpha }+\mathrm{i\beta }$is not a root of the associated auxiliary equation.

Comparing equations (1) and (2), we get:

M=0 and $\mathrm{\alpha }=0;\mathrm{\beta }=1$

Therefore, $\mathbf{\alpha }\mathbf{+}\mathbf{i\beta }\mathbf{=}\mathbf{i}$is not a root of the associated auxiliary equation so here s = 0

Assume, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{t}}^{\mathrm{s}}({\mathrm{A}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{\mathrm{\alpha x}}\mathrm{cos\beta x}+{\mathrm{t}}^{\mathrm{s}}({\mathrm{B}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{B}}_1\mathrm{t}+{\mathrm{B}}_0){\mathrm{e}}^{\mathrm{\alpha x}}\mathrm{sin\beta x}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)={\mathrm{\theta }}^0\left({\mathrm{A}}_0\right){\mathrm{e}}^{\left(0\right)\mathrm{\theta }}\cos \left(1\right)\mathrm{\theta }+{\mathrm{\theta }}^0\left({\mathrm{B}}_0\right){\mathrm{e}}^{\left(0\right)\mathrm{\theta }}\sin \left(1\right)\mathrm{\theta }\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)={\mathrm{A}}_0\mathrm{cos\theta }+{\mathrm{B}}_0\mathrm{sin\theta }\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)=\mathrm{Acos\theta }+\mathrm{Bsin\theta }......\left(3\right)\end{array}$

Now find the first and second derivatives of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{\theta }\right)=-\mathrm{Asin\theta }+\mathrm{Bcos\theta }\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{\theta }\right)=-\mathrm{Acos\theta }-\mathrm{Bsin\theta }\end{array}$

Substitute the value of ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)$and${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{\theta }\right)$the equation (1),

$\begin{array}{c}\mathrm{y}\text{'}\text{'}+4\mathrm{y}=\mathrm{sin\theta }-\mathrm{cos\theta }\\ -\mathrm{Acos\theta }-\mathrm{Bsin\theta }+4(\mathrm{Acos\theta }+\mathrm{Bsin\theta })=\mathrm{sin\theta }-\mathrm{cos\theta }\\ 3\mathrm{Acos\theta }+3\mathrm{Bsin\theta }=\mathrm{sin\theta }-\mathrm{cos\theta }\end{array}$

Comparing all coefficients of the above equation,

role="math" localid="1654947115962" $\begin{array}{c}3\mathrm{A}=-1\Rightarrow \mathrm{A}=\frac{-1}{3}\\ 3\mathrm{B}=1\Rightarrow \mathrm{B}=\frac{1}{3}\end{array}$

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

role="math" localid="1654947233973" $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)=\mathrm{Acos\theta }+\mathrm{Bsin\theta }\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)=\left(-\frac{1}{3}\right)\mathrm{cos\theta }+\left(\frac{1}{3}\right)\mathrm{sin\theta }\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)=\frac{1}{3}\mathrm{sin\theta }-\frac{1}{3}\mathrm{cos\theta }\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)=\frac{1}{3}(\mathrm{sin\theta }-\mathrm{cos\theta })\end{array}$

Therefore, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)=\frac{1}{3}(\mathrm{sin\theta }-\mathrm{cos\theta })$

Therefore, the general solution is,

$\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{\theta }\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)\\ \mathrm{y}={\mathrm{c}}_1\cos \left(2\mathrm{\theta }\right)+{\mathrm{c}}_2\sin \left(2\mathrm{\theta }\right)+\frac{1}{3}(\mathrm{sin\theta }-\mathrm{cos\theta })\end{array}$