Q4E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Verify that $\mathit{y}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathbf{3}\mathit{t}\mathbf{+}\mathbf{2}\mathit{c}\mathit{o}\mathit{s}\mathbf{3}\mathit{t}$ is a solution to the initial value problem

$\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{18}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}$

Find the maximum of $\left|y\left(t\right)\right|$ for $\mathbf{-}\mathbf{\infty }\mathbf{<}\mathit{t}\mathbf{<}\mathbf{\infty }$.

Step-by-Step Solution

Answer

Here $y=sin3t+2cos3t$ is a solution for the given differential equation, the maximum value $\left|y\right|=2.236$.

Step 1: Differentiate the value of y

To verify that $y=sin3t+2cos3t$ is a solution of $2y\text{'}\text{'}+18y=0$. 

 

Find $y\text{'}$ and $y\text{'}\text{'}.$

$$\begin{gathered} y\text{'}\left(t\right)=3cos3t-6sin3t \\ y\text{'}\text{'}\left(t\right)=-9sin3t-18cos3t \end{gathered}$$

 

Substitute these two equations in the differential equation;

$$\begin{gathered} 2(-9sin3t-18cos3t)+18(sin3t+2cos3t)=0 \\ -18sin3t-36cos3t+18sin3t+36cos3t=0 \\ 0=0 \end{gathered}$$ 

Therefore, LHS = RHS.

Step 2: Verify boundary condition.

Now verify the boundary conditions;

$$\begin{gathered} y\left(0\right)=sin(3\times 0)+2cos(3\times 0) \\ =2 \end{gathered}$$ 

And

$$\begin{gathered} y\text{'}\left(0\right)=3cos(3\times 0)-6sin(3\times 0) \\ =3 \end{gathered}$$ 

 

Therefore, $y=sin3t+2cos3t$ is a solution for the given differential equation.

Now plot the graph $y=sin3t+2cos3t$ and from the graph, one can see the maximum value $\left|y\right|=2.236$.