# Exact differential equation

## Homework Statement

Determine whether exact. If yes, solve.

(4t3y-15t2-y)dt + (t4+3y2-t)dy = 0

## The Attempt at a Solution

M= (4t3y-15t2-y)
N= (t4+3y2-t)

My=4t3-1
Nt= 4t3-1

So, yes it is exact.

fy= M = (4t3y-15t2-y)
f(t,y) = ∫ N dy
= 2t3y2-15t2y-(1/2)y2+g(t)

ft= 6t2y2-30ty+ g'(t)

This is where I've gotten stuck. I know I need to set this equal to M, and then all of the t's should cancel, but from what I've done, that won't work. So, this means I've made a mistake somewhere else.

Thank you!

LeonhardEuler
Gold Member
fy= M = (4t3y-15t2-y)
Then you fix it here:
f(t,y) = ∫ N dy
and then go back to the mistake:
= 2t3y2-15t2y-(1/2)y2+g(t)

tiny-tim
Homework Helper
hi musicmar!
f(t,y) = ∫ N dy
= 2t3y2-15t2y-(1/2)y2+g(t)

nooo, that's ∫ M dy, isn't it?

you need ∫ N dy

Yes, you're right. I just did it with N instead of M (oops.) and it worked out perfectly. Thank you so much!