1. ## Euler Differential Equation

Ok I have the equation:

theta^2(d2R/dtheta2)+3theta(dR/dtheta)-8R=20(theta^-3)

The condition is that R=10 when theta=1

I have worked out the equation as far as the General Solution.

The C.F i got was Ae^2t + Be^-4t

I then got a General Solution of Ae^2t + Be^-4t - 4e^-3t

How do I find the values of A and B ?

2. Originally Posted by LooNiE
Ok I have the equation:

theta^2(d2R/dtheta2)+3theta(dR/dtheta)-8R=20(theta^-3)

The condition is that R=10 when theta=1

I have worked out the equation as far as the General Solution.

The C.F i got was Ae^2t + Be^-4t

I then got a General Solution of Ae^2t + Be^-4t - 4e^-3t

How do I find the values of A and B ?
You have the condition $\theta (1) = 10$.
Where is your condition $\theta ' (1) = ?$.

3. Originally Posted by ThePerfectHacker
You have the condition $\theta (1) = 10$.
Where is your condition $\theta ' (1) = ?$.
R is finite as theta tends to infinity.

4. Originally Posted by LooNiE
R is finite as theta tends to infinity.
Okay. You said you got the general solution of $Ae^{2\theta} + Be^{-4\theta} - 4e^{-3\theta}$.
For this to stay bounded as $\theta \to \infty$ you need $A=0$.
Do you see why?

5. Originally Posted by ThePerfectHacker
Okay. You said you got the general solution of $Ae^{2\theta} + Be^{-4\theta} - 4e^{-3\theta}$.
For this to stay bounded as $\theta \to \infty$ you need $A=0$.
Do you see why?
yes i see. but how do i find B?

6. Originally Posted by LooNiE
yes i see. but how do i find B?
Once you let $A=0$ you are let with $Be^{-4\theta} - 4e^{-3\theta}$.
You are told that this expression is equal to $10$ with $\theta = 1$.
Thus, $Be^{-4} - 4e^{-3} = 10$.

Now solve for $B$.