Q7E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Question: Find a synchronous solution of the form $\mathbf{Acos\Omega t}\mathbf{+}\mathbf{Bsin\Omega t}$ to the given forced oscillator equation using the method of Example   to solve for  A and B $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{5}\mathbf{sin}\mathbf{3}\mathbf{t}\mathbf{,}\mathbf{\Omega }\mathbf{=}\mathbf{3}$.

 .

Step-by-Step Solution

Answer

Answer

 

The synchronous solution is for the oscillator equation .

Step 1: Finding the differential equation of Y

Given differential equation is  and $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{5}\mathbf{sin}\mathbf{3}\mathbf{t}$ 

 

The synchronous solution of the form $\mathbf{y}\mathbf{=}\mathbf{Acos}\mathbf{3}\mathbf{t}\mathbf{+}\mathbf{Bsin}\mathbf{3}\mathbf{t}.$ 

 

Substitute the value of Y  in the differential equation.

 $$\begin{gathered} \mathbf{y}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{Asin}\mathbf{3}\mathbf{t}\mathbf{+}\mathbf{3}\mathbf{Bcos}\mathbf{3}\mathbf{t}\mathbf{\dots }\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ \mathbf{y}\text{'}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{9}\mathbf{Acos}\mathbf{3}\mathbf{t}\mathbf{-}\mathbf{9}\mathbf{Bsin}\mathbf{3}\mathbf{t}\mathbf{\dots }\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

 

 

Step 2: Substitute the y ' and y '' in the given equation.

Substitute   and  in the differential equation;

 $$\begin{gathered} \mathbf{(}\mathbf{-}\mathbf{3}\mathbf{Asin}\mathbf{3}\mathbf{t}\mathbf{+}\mathbf{3}\mathbf{Bcos}\mathbf{3}\mathbf{t}\mathbf{)}\mathbf{+}\mathbf{4}\mathbf{(}\mathbf{Acos}\mathbf{3}\mathbf{t}\mathbf{+}\mathbf{Bsin}\mathbf{3}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{5}\mathbf{sin}\mathbf{3}\mathbf{t} \\ \mathbf{-}\mathbf{5}\mathbf{A}\mathbf{+}\mathbf{6}\mathbf{B}\mathbf{)}\mathbf{cos}\mathbf{3}\mathbf{t}\mathbf{+}\mathbf{(}\mathbf{-}\mathbf{6}\mathbf{A}\mathbf{-}\mathbf{5}\mathbf{B}\mathbf{)}\mathbf{sin}\mathbf{3}\mathbf{t}\mathbf{=}\mathbf{5}\mathbf{sin}\mathbf{3}\mathbf{t} \end{gathered}$$

 

 

Equate the coefficients of   and ;

 $$\begin{gathered} \mathbf{-}\mathbf{5}\mathbf{A}\mathbf{+}\mathbf{6}\mathbf{B}\mathbf{=}\mathbf{0}\mathbf{\dots }\mathbf{\left(}\mathbf{3}\mathbf{\right)} \\ \mathbf{-}\mathbf{6}\mathbf{A}\mathbf{-}\mathbf{5}\mathbf{B}\mathbf{=}\mathbf{5}\mathbf{\dots }\mathbf{\left(}\mathbf{4}\mathbf{\right)} \end{gathered}$$

 

Step 3: Finding the value of A

Multiply (3) with (4)  and (5) with   and then add two equations;

 $$\begin{gathered} \mathbf{-}\mathbf{25}\mathbf{A}\mathbf{+}\mathbf{30}\mathbf{B}\mathbf{=}\mathbf{0} \\ \mathbf{-}\mathbf{36}\mathbf{A}\mathbf{-}\mathbf{30}\mathbf{B}\mathbf{=}\mathbf{30} \\ \mathbf{61}\mathbf{A}\mathbf{=}\mathbf{30} \\ \mathbf{A}\mathbf{=}\mathbf{-}\frac{\mathbf{30}}{\mathbf{61}} \end{gathered}$$

 

Step 4: Finding the value of B

Substitute A  in  (3)

 $$\begin{gathered} \mathbf{-}\mathbf{5}\left(\mathbf{-}\frac{\mathbf{30}}{\mathbf{61}}\right)\mathbf{+}\mathbf{6}\mathbf{B}\mathbf{=}\mathbf{0} \\ \frac{\mathbf{150}}{\mathbf{61}}\mathbf{+}\mathbf{6}\mathbf{B}\mathbf{=}\mathbf{0} \\ \mathbf{6}\mathbf{B}\mathbf{=}\mathbf{-}\frac{\mathbf{150}}{\mathbf{61}} \\ \mathbf{B}\mathbf{=}\mathbf{-}\frac{\mathbf{25}}{\mathbf{61}} \end{gathered}$$

Therefore, the solution is .