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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Short 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-by-step solution

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;

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

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