How to solve differential equation problem?

Question

How to you solve the initial-value problem

a) y"+y = 0, y(0) = -1, y'(0) = -sqrt(3)
b) y"+9y = 0, y(0) = -1, y'(0) = sqrt(3)

Answer 1

Easy Homework.

I'll do it because you will be hopelessly lost if you don't get this.

First find the general solutions:
a) y=Asin(x) +Bcos(x)
b) y=Asin(3x)+Bcos(3x)

Plug in the initial conditions: you need to find y' to do this

a) y'=Acos(x)-Bsin(x). y(0)=-1 means -1=A*0+B*1, so B=-1.
y'(0)=-sqrt(3) means A*1-B*0=-sqrt(3), so A=-sqrt(3). So the solution is
y=-sqrt(3)*sin(x)-cos(x).

b) y'=3Acos(3x)-3Bsin(3x). Now, y(0)=-1 means A*0+B*1=-1, so B=-1. Also, y'(0)=-sqrt(3) means 3A*1-3B*0=-sqrt(3), so A=-sqrt(3)/3. Thus
y=(-sqrt(3)/3) sin(3x)-cos(3x).

That wasn't too hard, now was it?

Answer 2

These are homogenous linear ordinary differential equations, so they're pretty easy to solve. Let's handle (a) and you can try (b).

Since this is a *homogenous* *linear* ODE, assume that a solution to the problem (without initial values) is:

y = e^( r * t )

where r is some complex number. In that case,

y'' = r^2 * e^(r*t)

In that case, we can rewrite (a) as:

r^2 * e^(r*t) + e^(r*t) = 0

Of course, e^(r*t)>0 everywhere, so you can divide it out. In that case, you're left with (this is known as the "characteristic equation"):

r^2 + 1 = 0

Clearly, the answer to this is either r=-i or r=i where i=sqrt(-1). Now, you know that:

e^(i*t) = cos(t) + i*sin(t)

and

e^(-i*t) = cos(t) - i*sin(t)

Now, without getting into the details (this is a standard step; there is no need to explain this jump), this tells us that all solutions to the problem must take the form:

y(t) = A*cos(t) + B*sin(t)

where A and B are two real constants. Now, you might have wanted to start at this point with assuming this form of the solution. However, if your characteristic equation would have had real roots then the form of the solution would have been different (it would have been a real exponential) so it's important to start these problems with the CHARACTERISTIC EQUATION.

Now, this is where your initial conditions come in. If the solution takes that form, then:

y' = -A*sin(t) + B*cos(t)

So now you can plug in your initial conditions:

y(0) = A*cos(0) + B*sin(0) = A
y'(0) = -A*sin(0) + B*cos(0) = B

However, you know that y(0)=-1 and y'(0)=-sqrt(3), so A=-1 and B=-sqrt(3). So your solution (to (a)) is:

y(t) = -cos(t) - sqrt(3)*sin(t)

Follow the same procedure for part (b).