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!