Solve the initial value problem:
$$y'' + 3y = 0$$
$$y(0) = 1 , y'(0)= 3$$
$$r^2 + 3 = 0 $$
$$r = \pm i √3$$

Using Euler's equation:
$$\alpha = -b/(2a) , \beta = \sqrt{4ac-b^2}$$
$$\large e^{\alpha x}(c_1\cos \beta x+c_2\sin \beta x)$$

$$\alpha = 0 , \beta = \sqrt{3}$$

The general solution is:
$$y(x) = c_1\cos \sqrt{3}~x +c_2\sin \sqrt{3}~x$$
Am i right so far?..

Question

Solve the initial value problem: 
$$y'' + 3y = 0$$
$$y(0) = 1 , y'(0)= 3$$
$$r^2 + 3 = 0 $$
$$r = \pm i √3$$

Using Euler's equation:
$$\alpha = -b/(2a) , \beta = \sqrt{4ac-b^2}$$
$$\large e^{\alpha x}(c_1\cos \beta x+c_2\sin \beta x)$$

$$\alpha = 0 , \beta = \sqrt{3}$$

The general solution is:
$$y(x) = c_1\cos \sqrt{3}~x +c_2\sin \sqrt{3}~x$$
Am i right so far?..

shamil98 · Answer

$$y'(x) = -c_1 \sin \sqrt{3}~x + c_2 \cos \sqrt{3}~x$$

$$ 1 = c_1$$
$$3 = c_2$$
So:
$$y(x) = \cos \sqrt{3}~x +3\sin \sqrt{3}~x$$

math&ing001 · Answer

You forgot the sqrt(3) on the derivative, but other than that I don't see any mistakes.

shamil98 · Answer

Where on the derivative?..

math&ing001 · Answer

y'(x)=-c1*sqrt(3)*sin (sqrt(3)x) + c2*sqrt(3)*cos(sqrt(3)x)

shamil98 · Answer

I could rewrite it like this if that's what you meant.
$$y(x) = c_1\cos x\sqrt{3} +c_2\sin x\sqrt{3}$$

$$y'(x) = -c_1 \sin x\sqrt{3}+ c_2 \cos x\sqrt{3}$$
oh product rule, right, forgot about that .