# Thread: Simple Differential Equations Question

1. ## 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?

2. 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

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

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

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

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

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

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

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

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

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

Do you have a boundary condition so that you can find the unknown?

3. it says the point of origin

4. Then substitute $\displaystyle z = 0$ and $\displaystyle t = 0$ and solve for $\displaystyle C$.

5. 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?

6. 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...

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

Substitute the values

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

$\displaystyle 1 = C$.

So that means

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

I just realised too that you don't need the modulus signs at all...

7. 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.