Question

Find a particular solution to the differential equation.$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{9}$

Short answer

Thus, the particular solution to the differential equation is${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=-3.$

Step-by-step solution

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

The given differential equation,

$\mathrm{y}\text{'}\text{'}+3\mathrm{y}=-9\dots \left(1\right)$

The auxiliary equation for the above equation,

${\mathrm{m}}^2+3=0$

Now find the roots of the auxiliary equation.

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^2+3=0\\ {\mathrm{m}}^2=-3\\ \mathrm{m}=\pm \sqrt{-3}\\ \mathrm{m}=\pm \mathrm{i}\sqrt{3}\end{array}$

The roots of the auxiliary equation are:

${\mathrm{m}}_1=\mathrm{i}\sqrt{3}\&{\mathrm{m}}_2=-\mathrm{i}\sqrt{3}$

The complementary solution of the given equation is;

$\begin{array}{c}{\mathrm{y}}_{\mathrm{c}}\left(\mathrm{x}\right)=e^{0\cdot t}[{\mathrm{c}}_1\cos \left(\sqrt{3}\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(\sqrt{3}\mathrm{t}\right)]\\ ={\mathrm{c}}_1\cos \left(\sqrt{3}\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(\sqrt{3}\mathrm{t}\right)\end{array}$

Final conclusion, find a particular solution to the differential equation. 

Assume, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\mathrm{A}\dots \left(2\right)$

Now find the derivative of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{x}\right)=0\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{x}\right)=0\end{array}$

From the equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}+3{\mathrm{y}}_{\mathrm{p}}=-9\\ \left(0\right)+3\mathrm{A}=-9\\ \mathrm{A}=-3\end{array}$

Substitute the value of A in the equation (2), and we get:

$y_p\left(x\right)=-3$

Therefore, the particular solution of the given differential equation is:

$y_p\left(x\right)=-3$