MHB 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.
$$(2xy^2 + 2y) + (2x^2y +2x)y' = 0$$

So here is what I have so far..
$$M_y = 4xy+2$$
$$N_x = 4xy+2$$

so the equation is exact bc $$M_y = N_x$$

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

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

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

 

[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:

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

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

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

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

$$\pd{F}{x}=xy^2+y$$

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

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

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

$$\pd{F}{y}=x^2y+x$$

Hence, we obtain:

$$x^2y+x=x^2y+x+g'(y)$$

$$g'(y)=0$$

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

$$\frac{1}{2}x^2y^2+xy=C$$

or:

$$x^2y^2+2xy=C$$

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

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

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

 

[shamieh]

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

 

[Prove It]

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

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