# Simple Differential Equations Question

• Apr 1st 2010, 04:55 PM
proski117
Simple Differential Equations Question
I need help finding the solution to this differential equation.

dz/dt = 9te^(4z)

I cross multiply so that it becomes: dz/e^(4z) = 9t dt

From here I need to integrate.

on the left side, I do substitution and this is what I did:

u = e^(4z)
du = 4e^(4z) dz
du/4e^(4z) = dz

From here, my equation looked like:

du/4ue^(4z) = (9t^2)/2 + C
1/4e^(4z) ln|u| = (9t^2)/2 + C

I subbed in back the u in the absolute brackets but I am unsure of what to do next. I need to sub the points in f(0) = 0 but if I do using my equation, I am going to get 0, which I think is incorrect. Any ideas of where I went wrong?
• Apr 1st 2010, 05:38 PM
Prove It
Quote:

Originally Posted by proski117
I need help finding the solution to this differential equation.

dz/dt = 9te^(4z)

I cross multiply so that it becomes: dz/e^(4z) = 9t dt

From here I need to integrate.

on the left side, I do substitution and this is what I did:

u = e^(4z)
du = 4e^(4z) dz
du/4e^(4z) = dz

From here, my equation looked like:

du/4ue^(4z) = (9t^2)/2 + C
1/4e^(4z) ln|u| = (9t^2)/2 + C

I subbed in back the u in the absolute brackets but I am unsure of what to do next. I need to sub the points in f(0) = 0 but if I do using my equation, I am going to get 0, which I think is incorrect. Any ideas of where I went wrong?

There is no such thing as "cross multiplying" with derivatives. A derivative is NOT a fraction.

Anyway

$\frac{dz}{dt} = 9t\,e^{4z}$

$e^{-4z}\,\frac{dz}{dt} = 9t$

$\int{e^{-4z}\,\frac{dz}{dt}\,dt} = \int{9t\,dt}$

$\int{e^{-4z}\,dz} = \frac{9t^2}{2} + C_1$

$-\frac{1}{4}e^{-4z} + C_2 = \frac{9t^2}{2} + C_1$

$-\frac{1}{4}e^{-4z} = \frac{9t^2}{2} + C_1 - C_2$

$e^{-4z} = -18t^2 + C$ where $C = -4(C_1 - C_2)$

$-4z = \ln{|C - 18t^2|}$

$z = -\frac{1}{4}\ln{|C - 18t^2|}$.

Do you have a boundary condition so that you can find the unknown?
• Apr 1st 2010, 05:47 PM
proski117
it says the point of origin
• Apr 1st 2010, 06:05 PM
Prove It
Then substitute $z = 0$ and $t = 0$ and solve for $C$.
• Apr 1st 2010, 06:37 PM
proski117
by subbing in z,t = 0, you are left with:

-1/4 ln|c| = 0

what do you do from here? is your c = -1/4?
• Apr 1st 2010, 06:50 PM
Prove It
Quote:

Originally Posted by proski117
by subbing in z,t = 0, you are left with:

-1/4 ln|c| = 0

what do you do from here? is your c = -1/4?

No.

Go to the step that doesn't have a modulus sign in it...

$e^{-4z} = -18t^2 + C$

Substitute the values

$e^{-4(0)} = -18(0)^2 + C$

$1 = C$.

So that means

$z = -\frac{1}{4}\ln{(1 - 18t^2)}$.

I just realised too that you don't need the modulus signs at all...
• Apr 1st 2010, 06:52 PM
proski117
it says i am trying to look for the z value, and i entered -1/4 and it was incorrect

Edit: I think i am supposed to enter the equation, not the value of C. Thank you very much for your help.