Is this Differential Equation Exact?

[musicmar]

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)

M_y=4t³-1
N_t= 4t³-1

So, yes it is exact.


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

f_t= 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]

Here's your problem:

f_y= 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]

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

 

[musicmar]

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