# Finding a solution at a point

1. Nov 4, 2006

### hbomb

Give the general solution of the set of equations below:

x'=5x-2y+3z
y'=y
z'=6x+7y-2z

Which I found to be:
x=3C1e^7t-3C2e^t-C3e^-4t
y=6C2e^t
z=2C1e^7+8C2e^t+3C3e^-4t

Here's where I'm stuck. They want me to find the solution at t=0, (x,y,z)=(2,3,-1)

Which the professor hasn't told us how to do this.

2. Nov 4, 2006

### Office_Shredder

Staff Emeritus
I'm sure you know how to do this. If t=0, what is x(0)? There are two ways to express this... one is from the equation for x you solved for, the other is from the point you were given. Since t=0, all those nasty exponentials go away, and you have a system of three variables with constant coefficients (which is remarkably simpler than the initial set of differential equations you solved)

3. Nov 4, 2006

### hbomb

ok, so what I understand from this is that I plug t=0 into all the t's of the general solutions.

x=3C1-32-C3
y=6C2
z=2C1+8C2+3C3

And then do I set these equal to (x,y,z)=(2,3,-1). What is the solutions form suppose to look like?

4. Nov 4, 2006

### Office_Shredder

Staff Emeritus
Let's call $$C_1 C_2$$ and $$C_3$$ A, B, and C to make it easier.

You know
x(0)=3A - 3B - C=2
y(0)=6B = 3
z(0)=2A + 8B + 3C=-1

So you should be able to solve for A, B and C as numbers. For example, if 6B=3, B=2. Now you know 3A-6-C = 2, and 2A + 16 + 3C = -1