Second Order Linear Differential Equation Question

[VeganGirl]

Homework Statement

Solve the IVP, \frac{1}{4}y'' + 16y = 0
y(0)=\frac{1}{4}
y'(0)=0

Answer is given... y(t) = \frac{1}{4}cos 8t

Homework Equations

The Attempt at a Solution

This has the characteristic equation \frac{1}{4} \lambda^2 +16\lambda=0
Solving for lambda, I got \lambda= 0 or -64

Therefore y(t) = A*e^(0t) + B*e^(-64t) for some constants A and B
\Rightarrow y(t) = A + B*e^(-64t)

I know that I'll have to impose the initial conditions to get the specific solution, but my general solution is very different from the answer given. What am I doing wrong?

 

[rock.freak667]

The Attempt at a Solution

This has the characteristic equation \frac{1}{4} \lambda^2 +16\lambda=0

This part is wrong. If you use the solution y =e^rt, your equation will be

(1/4)r²+16=0

which will give you complex roots.

 

[Mark44]

Homework Statement

Solve the IVP, \frac{1}{4}y'' + 16y = 0
y(0)=\frac{1}{4}
y'(0)=0

Answer is given... y(t) = \frac{1}{4}cos 8t

Homework Equations

The Attempt at a Solution

This has the characteristic equation \frac{1}{4} \lambda^2 +16\lambda=0

No, the characteristic equation is
\frac{1}{4}\lambda^2 +16 = 0

or, equivalently,
\lambda^2 +64 = 0

Solving for lambda, I got \lambda= 0 or -64

Therefore y(t) = A*e^(0t) + B*e^(-64t) for some constants A and B
\Rightarrow y(t) = A + B*e^(-64t)

I know that I'll have to impose the initial conditions to get the specific solution, but my general solution is very different from the answer given. What am I doing wrong?

 

[VeganGirl]

Ohhh that's where I went wrong. Thanks guys!