How Do You Calculate Specific Values for Constants in Differential Equations?

  • Thread starter Thread starter hbomb
  • Start date Start date
  • Tags Tags
    Point
hbomb
Messages
57
Reaction score
0
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.
 
Physics news on Phys.org
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)
 
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?
 
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
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top