Q24E

Solve the given initial value problem. $y'' + 9 y = 0;$ $y (0) = 1,$ $y' (0) = 1$

Verified

Answer

The solution of the given initial value $y'' + 9 y = 0$ is $y (t) = \cos 3 t + \frac{}{} \sin 3 t$ when $y (0) = 1$ and $y' (0) = 1$ .

1Step 1: Complex conjugate roots.

If the auxiliary equation has complex conjugate roots $α \pm iβ$ , then the general solution is given as:

$y (t) = c_{1} e^{αt} cosβt + c_{2} e^{αt} sinβt$

2Step 2: Finding the roots of the auxiliary equation.

Given differential equation is $y'' + 9 y = 0$

Then the auxiliary equation is;

$\begin{matrix} r^{2} + 9 = 0 \\ r^{2} = - 9 \\ r = \pm \sqrt{- 9} \\ r = \pm 3 i \end{matrix}$

Therefore, the general solution is:

$\begin{array}{l} y (t) = e^{0 \times t} (c_{1} \cos (3 t) + c_{2} \sin (3 t)) \\ = c_{1} \cos (3 t) + c_{2} \sin (3 t)) \end{array}$

3Step 3: Finding the values of c 1 and c 2

Given initial conditions are $y (0) = 1$ and $y' (0) = 1$

$\begin{array}{l} y (0) = (c_{1} \cos (3 \times 0) + c_{2} \sin (3 \times 0)) \\ c_{1} = 1 \end{array}$

And $y' (t) = (- c_{1} \sin (3 t) + c_{2} \cos (3 t))$

Then $y' (0) = (- 3 c_{1} \sin (3 \times 0) + 3 c_{2} \cos (3 \times 0))$

$\begin{matrix} 3 c_{2} = 1 \\ c_{2} = \frac{1}{3} \end{matrix}$

Therefore, the solution is $y (t) = \cos 3 t + \frac{1}{3} \sin 3 t$