Q9E - 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 4  to solve for A  and B$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{6}\mathbf{cos}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{8}\mathbf{sin}\mathbf{2}\mathbf{t}\mathbf{,}\mathbf{\Omega }\mathbf{=}\mathbf{2}$  .

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Step-by-Step Solution

Answer

Answer

 

The synchronous solution to the given forced oscillator equation is $\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{cos}\mathbf{2}\mathbf{t}\mathbf{+}\frac{\mathbf{3}}{\mathbf{2}}\mathbf{sin}\mathbf{2}\mathbf{t}.$

Step 1: Finding the differential equation of Y

Given differential equation is  $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{6}\mathbf{cos}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{8}\mathbf{sin}\mathbf{2}\mathbf{t}\mathbf{,}\mathbf{\Omega }\mathbf{=}\mathbf{2}.$

 

The synchronous solution of the form $\mathbf{y}\mathbf{=}\mathbf{Acos}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{Bsin}\mathbf{2}\mathbf{t}.$  

 

Now substitute the value of y in the differential equation;

 $$\begin{gathered} \mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{Asin}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{2}\mathbf{Bcos}\mathbf{2}\mathbf{t}\mathbf{\dots }\left(\mathbf{1}\right) \\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{Acos}\mathbf{2}\mathbf{t}\mathbf{-}\mathbf{4}\mathbf{Bsin}\mathbf{2}\mathbf{t}\mathbf{\dots }\left(\mathbf{2}\right) \end{gathered}$$

 

 

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

Substitute   and   in the differential equation;

 

 $$\begin{gathered} \mathbf{-}\mathbf{4}\mathbf{Acos}\mathbf{2}\mathbf{t}\mathbf{-}\mathbf{4}\mathbf{Bsin}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{2}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{Asin}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{2}\mathbf{Bcos}\mathbf{2}\mathbf{t}\mathbf{)}\mathbf{+}\mathbf{4}\mathbf{(}\mathbf{Acos}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{Bsin}\mathbf{2}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{6}\mathbf{cos}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{8}\mathbf{sin}\mathbf{2}\mathbf{t}\mathbf{4}\mathbf{Bcos}\mathbf{2}\mathbf{t}\mathbf{-}\mathbf{4}\mathbf{Asin}\mathbf{2}\mathbf{t} \\ \mathbf{=}\mathbf{6}\mathbf{cos}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{8}\mathbf{sin}\mathbf{2}\mathbf{t} \end{gathered}$$

Step 3: Finding the value of B

Equate the coefficients of $\mathbf{cos}\mathbf{2}\mathbf{t}$, then$\mathbf{4}\mathbf{B}\mathbf{=}\mathbf{6}$ we get 

 $$\begin{gathered} \mathbf{B}\mathbf{=}\frac{\mathbf{6}}{\mathbf{4}} \\ \mathbf{=}\frac{\mathbf{3}}{\mathbf{2}} \end{gathered}$$

 

 

Step 4: Finding the value of A

Equate the coefficients of  sin2t , then we get;

 $$\begin{gathered} \mathbf{-}\mathbf{4}\mathbf{A}\mathbf{=}\mathbf{8} \\ \mathbf{A}\mathbf{=}\mathbf{-}\frac{\mathbf{8}}{\mathbf{4}} \\ \mathbf{A}\mathbf{=}\mathbf{-}\mathbf{2} \end{gathered}$$

 

Therefore, the solution is $\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{cos}\mathbf{2}\mathbf{t}\mathbf{+}\frac{\mathbf{3}}{\mathbf{2}}\mathbf{sin}\mathbf{2}\mathbf{t}$.