Exact Differential Equations: Solving for F(t,y)

[jumbogala]

Homework Statement

(28y⁶ - 24e^-6y)(sqrt(1-t⁴))dy -(10t)dt =0

Come up with a solution to this equation in the form of a function F(t,y)=0.

Homework Equations

dF/ dy = N
dF / dt = M

Note: when I write d it's supposed to be a partial derivative.

The Attempt at a Solution

Okay, so I called the part in front of the dy 'N' and the part in front of the dt 'M'. Then I took dM/dy and dN/dt to make sure they're equal. They are, so this equation is exact.

F(t,y) = integral of (-10t)dt = -5t² + k(y)

Now, take the partial derivative of F(t,y) with respect to y. This gives -5t²y + k`(y). This equal to N.

So -5t²y + k`(y) = (28y⁶ - 24e^-6y)(sqrt(1-t⁴))

There's where I get stuck. I need to figure out what k(y) is, but I have nooo idea how to do that. I tried moving everything over to the right side except k`(y) and integrating, but the integral got really complicated =\

 

[206PiruBlood]

Remember that you have two variables so if you integrate with respect to one, you hold the other constant. For example if you integrate with respect to y, then you would treat t as a constant.

 

[jumbogala]

Hmm, okay. I think I see where I went wrong.

F = integral of M dt = -5t^2 + k(y)

Now, take the partial derivative of F with respect to y. This gives k`(y).

So k`(y) = (28y^6 - 24e^(-6y))(sqrt(1-t^4))

So to find k(y), I integrated the right hand side of the above equation with respect to y. Then I got k(y) = (sqrt(1-t^4))(4y^7 + 4e^(-6y)).

But my answer was still wrong...

 

[Mark44]

Why is it that you think this differential equation is exact? I'm not getting that at all. Based on your definitions for M and N,
$$\frac{\partial M}{\partial y} = 0$$
and
$$\frac{\partial N}{\partial t} = (28y^6 - 24e^{-6y})(1/2)(1 - t^4)^{-1/2}(-4t^3)$$
Are you sure you have written the problem down correctly?