Question

Find a particular solution to the differential equation.

$\mathbf{\theta }\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{-}\mathbf{\theta }\left(\mathbf{t}\right)\mathbf{=}\mathbf{tsint}$

Short answer

The particular solution of the differential equation is${\mathbf{\theta }}_p\mathbf{=}\mathbf{-}\frac{1}{2}\mathbf{tsin}\left(\mathbf{t}\right)\mathbf{-}\frac{1}{2}\mathbf{cos}\left(\mathbf{t}\right)$.

Step-by-step solution

Firstly, write the auxiliary equation of the above differential equation.

The differential equation;

$\mathrm{\theta }\text{'}\text{'}\left(\mathrm{t}\right)-\mathrm{\theta }\left(\mathrm{t}\right)=\mathrm{tsint}\dots \left(1\right)$

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

$\mathrm{\theta }\text{'}\text{'}\left(\mathrm{t}\right)-\mathrm{\theta }\left(\mathrm{t}\right)=0$

The auxiliary equation for the above equation,

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

Now find the roots of the auxiliary equation

Solve the auxiliary equation:

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

The roots of the auxiliary equation are:

${\mathrm{m}}_1=1,\&{\mathrm{m}}_2=-1$

The complementary solution of the given equation is:

${\mathrm{\theta }}_{\mathrm{c}}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{x}}$

Use the method of undetermined coefficients to find a particular solution to the differential equation. 

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

${\mathrm{\theta }}_{\mathrm{p}}=(\mathrm{At}+\mathrm{B})\sin \left(\mathrm{t}\right)+(\mathrm{Ct}+\mathrm{D})\cos \left(\mathrm{t}\right)......\left(2\right)$

Now find the derivative of the above equation,

$\begin{array}{c}{\mathrm{\theta }}_{\mathrm{p}}\text{'}=(\mathrm{At}+\mathrm{B})\mathrm{cost}+\mathrm{Asint}+(\mathrm{Ct}+\mathrm{D})(-\mathrm{sint})+\mathrm{Ccost}\\ {\mathrm{\theta }}_{\mathrm{p}}\text{'}\text{'}=(\mathrm{At}+\mathrm{B})(-\mathrm{sint})+\mathrm{Acost}+\mathrm{Acost}+(\mathrm{Ct}+\mathrm{D})(-\mathrm{cost})+\mathrm{C}(-\mathrm{sint})-\mathrm{Csint}\\ {\mathrm{\theta }}_{\mathrm{p}}\text{'}\text{'}=(\mathrm{At}+\mathrm{B})(-\mathrm{sint})+2\mathrm{Acost}+(\mathrm{Ct}+\mathrm{D})(-\mathrm{cost})-2\mathrm{C}\left(\mathrm{sint}\right)\end{array}$

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

$\begin{array}{c}{\mathrm{\theta }}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)-{\mathrm{\theta }}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{tsint}\\ (\mathrm{At}+\mathrm{B})(-\mathrm{sint})+2\mathrm{Acost}+(\mathrm{Ct}+\mathrm{D})(-\mathrm{cost})-2\mathrm{Csint}-[(\mathrm{At}+\mathrm{B})\mathrm{sint}+(\mathrm{Ct}+\mathrm{D})\mathrm{cost}]=\mathrm{tsint}\\ (-2\mathrm{A})\mathrm{tsint}+(-2\mathrm{B}-2\mathrm{C})\mathrm{sint}+(-2\mathrm{C})\mathrm{tcost}+(2\mathrm{A}-2\mathrm{D})\mathrm{cost}=\mathrm{tsint}\end{array}$

Final conclusion.

Comparing all coefficients of the above equation;

$\begin{array}{l}-2\mathrm{A}=1\Rightarrow \mathrm{A}=\frac{-1}{2}\\ -2\mathrm{C}=0\Rightarrow \mathrm{C}=0\\ -2\mathrm{B}-2\mathrm{C}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(3\right)\\ 2\mathrm{A}-2\mathrm{D}=0\dots \left(4\right)\end{array}$

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

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

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

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

Therefore, the particular solution of equation (1),