Separable Equation: dz/dt = 9t e^(6z) that passes through the origin

[CollegeGirl001]

Problem:
dz/dt = 9t e^(6z)
that passes through the origin

My attempt:

• -e^(6z) dz = 9t dt
• integrate both sides
• -1/6 e^(6z) = 9/2 t^2 + C Where C is a constant
• e^(6z) = -6 (9/2 t^2 + C)
• e^(6z) = ((-27 t^2) - 6C)

this step where I get a little doubtful...

• 6z = ln ((-27 t^2) - 6C)
• z = 1/6 ln ((-27 t^2) - 6C)

Now I need to solve for C. I know that z(0)=0

• 0= 1/6 ln (-6C)

I don't know what to do from here. Can you help please?

 

[stapel]

Problem:
dz/dt = 9t e^(6z)
that passes through the origin

My attempt:

• -e^(6z) dz = 9t dt

Did you really subtract the exponential in order to move it to the other side...?

 

[CollegeGirl001]

Then what do I do first?

 

[CollegeGirl001]

Thank you

ohhhhhhhhhhh, haha
Yes, I did subtract it.
I was up late working on my homework. I must have been extremely tired.
Thank you. I'll try working on it again.