# Find value of K for differential equation to be exact

1. Sep 27, 2015

### masterchiefo

1. The problem statement, all variables and given/known data
Hello everyone,
I need to find K for the following differential equation to be exact.
(y3+kxy4-2*x)dx +(3xy2 +20x2y3)dy=0

2. Relevant equations

3. The attempt at a solution
(y3+kxy4-2*x)dx +(3xy2 +20x2y3)dy=0

dM/dy = d/dy(y3+kxy4-2*x) = 4*k*y3*x+3*y2

dN/dx =d/dx(3xy2 +20x2y3) = 40*x*y3+3*y2

After that I really have no idea how to continue this.

thank you.

2. Sep 27, 2015

### davidmoore63@y

You have done 98% of it. What value of k makes the two expressions equal?

3. Sep 27, 2015

### Staff: Mentor

Your diff. equation will be exact if My = Nx; that is, if 4kxy3 + 3y2 = 40xy3 + 3y2. What does k have to be so that the two expressions are identically equal?

4. Sep 27, 2015

### masterchiefo

oh wow I am so sleepy.

K have to be 10.
But in this particular equation its pretty easy to spot out, but what if both part is very different, how do I proceed?

5. Sep 27, 2015

### Staff: Mentor

Yes, k = 10. If the equation isn't exact, there are several alternatives, but they are too involved to explain in a small space like this. If you read ahead in your book, they might have some discussion of these techniques.

6. Sep 27, 2015

### masterchiefo

thanks also, to find the solution of current equation after finding k=10

I do
integral(40*x*y3+3*y2)dx = 20*x2*y3+3*x*y2
integral(40*x*y3+3*y2)dy =10*y4*x+y3

I look for same terms in both results, but here I see that there isnt.

so c= 20*x2*y3+3*x*y2+ 10*y4*x+y3

is that correct ?

7. Sep 27, 2015

### Staff: Mentor

No. Integrate M with respect to x, keeping in mind that there might be a "constant" of integration that would be either a constant or a function of y alone, say g(y).
Then integrate N with respect to y, also keeping in mind that there could be a "constant" of integration that would be a constant or a function of x alone, say h(x).

What you're doing here is assuming that there is an equation F(x, y) = C. If you take the total derivative of both sides, you get
$$\frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y} dy = 0$$
Here, M = $\frac{\partial F}{\partial x}$ and N = $\frac{\partial F}{\partial y}$, and you're trying to reconstruct F from its two partial derivatives.

8. Sep 27, 2015

### masterchiefo

I don't understand.

M(x,y) = 40*x*y3+3*y2
N(x,y)= 40*x*y3+3*y2

I don't simply integrate each ?

9. Sep 27, 2015

### Staff: Mentor

From the original problem, M = y3 + 10xy4, and N = 3xy2 + 40xy3.

What you have above are My and Nx. The condition for exactness is that these two partials should be equal. To solve the diff. eqn, you need to do as I suggest in my previous post.

10. Sep 27, 2015

### masterchiefo

= integral(y3+10xy4-2*x)dx
= x2*(5*y4-1)+x*y3 +h(y)
= x2*5*y4-x2+x*y3 +h(y)
h(y) = n/a

integral(3xy2 +20x2y3)dy
= 5*y4*x2+y3*x +g(x)
g(x) = -x2

c= x2*5*y4-x2+x*y3

11. Sep 27, 2015

### Staff: Mentor

Yes, looks good.

To check, take the total differential of both sides. (I mistakely said total derivative earlier.)

Your solution F(x, y) = C, where $F(x, y) = 5x^2y^4 + xy^3 - x^2$
Taking the total differential of both sides, we get d(F(x, y)) = d(C)
or, $\frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy = 0$
With the partial derivatives calculated, you should get your original diff. equation.