Question

Find the solution of the given initial value problem:
y''+3y=0
y(0)= -2
y'(0)= 3

Answer 1

Well I figured it out if anyone was interested:
y=e^(mx) y'=me^(mx) y''=(m^2)e^(mx)

Knowing this, just plug it back into the equation:
(m^2)e^(mx)+3(e^(mx)) = 0

Divide by e^(mx) which gets you:
m^2 + 3 = 0

Solve for m:
m= sqrt(3) and m= -sqrt(3)

General solution:
y=C1e^sqrt(3) + C2e^-sqrt(3)

Use initial values:
-2 = C1 + C2
3 = sqrt(3)C1 - sqrt(3)C2

Solve for C1 and C2 by substitution and plug them back into the general solution!

Answer 2

You made a slight mistake in solving \(m^2+3=0\). The equation has two roots\(m_1=\sqrt{3}i\) and \(m_2=-\sqrt{3}i\). The solution will be then \(y(x)=c_1\cos(\sqrt{3}x)+c_2\sin(\sqrt{3}x)\). You can now find the constants by pluggin the given initial values.

Answer 3

If you're wondering how the solution came to be in terms of cosine and sine functions, then you should know that \(e^{ix}=\cos{x}+i\sin{x}\).