Determine whether the equation is exact. If it is exact, find the solution

[shamieh]

Determine whether the equation is exact. If it is exact, find the solution.
\(\displaystyle (2xy^2 + 2y) + (2x^2y +2x)y' = 0\)

So here is what I have so far..
\(\displaystyle M_y = 4xy+2\)
\(\displaystyle N_x = 4xy+2\)

so the equation is exact bc \(\displaystyle M_y = N_x\)

So I can't find the devil pitchfork and I forget what it is called lol..
but

\(\displaystyle Pitchfork_x = x^2y^2 + 2xy + h(y)\)
\(\displaystyle Pitchfork_y = h(y) = x^2y^2 + 2xy\)

so \(\displaystyle Pitchfork_{(x,y)} = 2x^2y^2 + 4xy = c\) ?

 

[MarkFL]

I'm not sure what you mean by "pitchfork," but this is the way I was taught to solve exact equations...

First, let's write the equation in differential form:

\(\displaystyle \left(2xy^2+2y\right)\,dx+\left(2x^2y+2x\right)\,dy=0\)

To make things simpler, let's divide through by 2:

\(\displaystyle \left(xy^2+y\right)\,dx+\left(x^2y+x\right)\,dy=0\)

By inspection we can see the equation is exact. Since the equation is exact, we must therefore have:

\(\displaystyle \pd{F}{x}=xy^2+y\)

We now integrate this with respect to $x$ to get:

\(\displaystyle F(x,y)=\frac{1}{2}x^2y^2+xy+g(y)\)

Now, to determine $g(y)$, we may take the partial derivatives with respect to $y$, and observe that we must have:

\(\displaystyle \pd{F}{y}=x^2y+x\)

Hence, we obtain:

\(\displaystyle x^2y+x=x^2y+x+g'(y)\)

\(\displaystyle g'(y)=0\)

Since $g$ is just a constant, we obtain::

\(\displaystyle \frac{1}{2}x^2y^2+xy=C\)

or:

\(\displaystyle x^2y^2+2xy=C\)

This is equivalent to the implicit solution you obtained. :D

From this we may obtain the explicit solution (by considering the quadratic in $xy$):

\(\displaystyle y(x)=\frac{c_1}{x}\)

 

[shamieh]

oh thanks. & yea I was talking about this thing \(\displaystyle \psi\) psi i guess its called after some extensive google research.

 

[Prove It]

oh thanks. & yea I was talking about this thing \(\displaystyle \psi\) psi i guess its called after some extensive google research.

More likely a capital Psi $\displaystyle \begin{align*} \Psi \end{align*}$

 

[blue_raver22]

I would first clarify that the equation being referred to is a differential equation, as it involves a derivative. I would also mention that finding the solution using the "pitchfork method" is a common approach in solving exact differential equations.

To answer the question, yes, the equation is exact since the partial derivatives of the given function with respect to x and y are equal. The solution can be found by first finding the integrating factor, which in this case is e^(x^2y^2). Multiplying both sides of the equation by this integrating factor, we get:

e^(x^2y^2)(2xy^2 + 2y) + e^(x^2y^2)(2x^2y +2x)y' = 0

Using the product rule, we can simplify the left side to:

(e^(x^2y^2)2xy^2)' + (e^(x^2y^2)2x^2y)'y' = 0

Integrating both sides with respect to y, we get:

e^(x^2y^2)2xy^2 + C(x) = 0

where C(x) is a constant of integration with respect to y. Solving for C(x), we get:

C(x) = -e^(x^2y^2)2xy^2

Therefore, the solution to the given differential equation is:

e^(x^2y^2)2xy^2 + e^(x^2y^2)2x^2y = c

where c is a constant of integration.