Find value of K for differential equation to be exact

[masterchiefo]

Homework Statement

Hello everyone,
I need to find K for the following differential equation to be exact.
(y³+kxy⁴-2*x)dx +(3xy² +20x²y³)dy=0

Homework Equations

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.

 

[davidmoore63@y]

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

 

[Mark44]

Homework Statement

Hello everyone,
I need to find K for the following differential equation to be exact.
(y³+kxy⁴-2*x)dx +(3xy² +20x²y³)dy=0

Homework Equations

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.

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

 

[masterchiefo]

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

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

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?

 

[Mark44]

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?

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.

 

[masterchiefo]

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.

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 ?

 

[Mark44]

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 ?

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.

 

[masterchiefo]

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).

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 ?

 

[Mark44]

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 ?

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

What you have above are M_y and N_x. 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.

 

[masterchiefo]

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

What you have above are M_y and N_x. 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.

= 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

 

[Mark44]

= 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

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.