• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q13E

Expert-verified
Found in: Page 180

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# Find a particular solution to the differential equation.${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{9}}{\mathbf{y}}{\mathbf{=}}{\mathbf{3}}{\mathbf{sin}}{\mathbf{3}}{\mathbf{t}}$

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

See the step by step solution

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

The differential equation is:

$\mathrm{y}\text{'}\text{'}-\mathrm{y}\text{'}+9\mathrm{y}=3\mathrm{sin}3\mathrm{t}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(1\right)$

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,

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

The roots of the auxiliary equation are,

${\mathrm{m}}_{1}=\frac{1+\mathrm{i}\sqrt{35}}{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}&\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{m}}_{2}=\frac{1-\mathrm{i}\sqrt{35}}{2}$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{e}}^{\frac{t}{2}}{\mathrm{c}}_{1}\mathrm{cos}\left(\frac{\sqrt{35}}{2}\mathrm{t}\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(\frac{\sqrt{35}}{2}\mathrm{t}\right)$

## 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),

${\mathrm{y}}_{\mathrm{p}}=\mathrm{Asin}\left(3\mathrm{t}\right)+\mathrm{Bcos}\left(3\mathrm{t}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(2\right)$

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 ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{\mathrm{p}}\text{'}$ and ${\mathrm{y}}_{\mathrm{p}}$ in the equation (1),

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

## 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),

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