Question

In Example 3, we used the second solution in (5.24), and obtained (5.25) as the particular solution satisfying the given initial conditions. Show that the first and third solutions in (5.24) also give the particular solution (5.25) satisfying the given initial conditions.

Short answer

The particular solution is$y=10\sin (\omega t+90).$

Step-by-step solution

 Step 1: Given information from question

The given initial conditions are $y=-10\text{and}dy/dx=0$, when.$t=0$

General solution

General solution of first firm is$y=A{e}^{i\omega t}+B{e}^{-i\omega t}$

Use the first firm of general solution and its derivative

Use the initial condition $y=-10$and $dy/dx=0$at time$t=0$. Using the first firm of general solutionand its derivative

$\begin{array}{c}y=A{e}^{i\omega t}+B{e}^{-i\omega t}\\ \frac{dy}{d\omega }=i\omega A{e}^{i\omega t}-i\omega B{e}^{-i\omega t}\end{array}$

When apply the initial condition

$\begin{array}{c}-10=A+B\\ 0=i\omega (A-B)\end{array}$

The value of constant and by above condition,

$\begin{array}{c}A=B=-\frac{10}{2}\\ =-5\end{array}$

Then the particular solution

$\begin{array}{c}y=\frac{10}{2}(e^{i\omega t}+e^{-i\omega t})\\ =-10\cos \omega t.\end{array}$

Apply the initial condition on the third solution

Apply the initial condition on the third solution and find its first derivative.

$\begin{array}{c}y=c\sin (\omega t+\gamma )\\ \frac{dy}{dx}=c\omega \cos (\omega t+\gamma )\end{array}$

Therefore, it gets

$\begin{array}{c}-10=c\sin \gamma \to c=\frac{-10}{\sin \gamma }\\ 0=c\sin \gamma \to \gamma =90\\ \gamma =90,c=-10\end{array}$

Therefore, the particular solution is$y=10\sin (\omega t+90).$