Q13E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a particular solution to the differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{3}\mathbf{sin}\mathbf{3}\mathbf{t}$

Step-by-Step Solution

Answer

The particular solution of the given differential equation is  ${\mathrm{y}}_{\mathrm{p}}=\cos \left(3\mathrm{t}\right).$

Step 1: Firstly, write the auxiliary equation of the above differential equation.

The differential equation is:

 $\mathrm{y}\text{'}\text{'}-\mathrm{y}\text{'}+9\mathrm{y}=3\sin 3\mathrm{t}......\left(1\right)$

 

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

 

$\mathrm{y}\text{'}\text{'}-\mathrm{y}\text{'}+9\mathrm{y}=0$

 

The auxiliary equation for the above equation,

 

${\mathrm{m}}^2-\mathrm{m}+9=0$

 

Step 2: Now find the roots of the auxiliary equation.

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^2-\mathrm{m}+9=0\\ \mathrm{m}=\frac{-(-1)\pm \sqrt{1-4\left(9\right)\left(1\right)}}{2}\\ \mathrm{m}=\frac{1\pm \sqrt{-35}}{2}\\ \mathrm{m}=\frac{1\pm \mathrm{i}\sqrt{35}}{2}\end{array}$

The roots of the auxiliary equation are, 

${\mathrm{m}}_1=\frac{1+\mathrm{i}\sqrt{35}}{2},\&{\mathrm{m}}_2=\frac{1-\mathrm{i}\sqrt{35}}{2}$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{e}}^{\frac{t}{2}}{\mathrm{c}}_1\cos \left(\frac{\sqrt{35}}{2}\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(\frac{\sqrt{35}}{2}\mathrm{t}\right)$

Step 3: 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),

 

${\mathrm{y}}_{\mathrm{p}}=\mathrm{Asin}\left(3\mathrm{t}\right)+\mathrm{Bcos}\left(3\mathrm{t}\right)\dots \left(2\right)$

 

Now find the derivative of the above equation,

 $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}=3\mathrm{Acos}\left(3\mathrm{t}\right)-3\mathrm{Bsin}\left(3\mathrm{t}\right)\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}=-9\mathrm{Asin}\left(3\mathrm{t}\right)-9\mathrm{Bcos}\left(3\mathrm{t}\right)\end{array}$

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

 $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}-{\mathrm{y}}_{\mathrm{p}}\text{'}+9{\mathrm{y}}_{\mathrm{p}}=3\sin 3\mathrm{t}\\ -9\mathrm{Asin}\left(3\mathrm{t}\right)-9\mathrm{Bcos}\left(3\mathrm{t}\right)-\left(3\mathrm{Acos}\left(3\mathrm{t}\right)\right)+9(+\mathrm{Bcos}\left(3\mathrm{t}\right))=3\sin 3\mathrm{t}\\ (-9\mathrm{A}+3\mathrm{B}+9\mathrm{A})\sin \left(3\mathrm{t}\right)+(-9\mathrm{B}-3\mathrm{A}+9\mathrm{B})\cos \left(3\mathrm{t}\right)=3\sin 3\mathrm{t}\\ 3\mathrm{Bsin}\left(3\mathrm{t}\right)-3\mathrm{Acos}\left(3\mathrm{t}\right)=3\sin 3\mathrm{t}\end{array}$

Step 4: Final conclusion.

Comparing all coefficients of the above equation;

 

$\begin{array}{c}3\mathrm{B}=3\\ \mathrm{B}=1\\ -3\mathrm{A}=0\\ \mathrm{A}=0\end{array}$

Therefore, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}=\mathrm{Asin}\left(3\mathrm{t}\right)+\mathrm{Bcos}\left(3\mathrm{t}\right)\\ {\mathrm{y}}_{\mathrm{p}}=\left(0\right)\sin \left(3\mathrm{t}\right)+\left(1\right)\cos \left(3\mathrm{t}\right)\\ {\mathrm{y}}_{\mathrm{p}}=\cos \left(3\mathrm{t}\right)\end{array}$