Solving a Non-Exact ODE: $(y-2x^2y)dx +xdy = 0$

[NotaMathPerson]

Solve the ode

$$(y-2x^2y)dx +xdy = 0$$

The equation is in exact form $$Q(x,y)dx+ P(x,y)dy =0$$

When I test for exactness it fails. Then I used the technique $$\frac{M_y-N_x}{N}$$

I get

$u(x)=-2x$ as my integrating factor.

But I end still end up with a non-exact d.e why is that?

$$(-2xy+4x^3y)dx-2x^2dy=0$$

 

[Ackbach]

It's separable. Factor out the $y$ on the $dx$ term.

 

[NotaMathPerson]

It's separable. Factor out the $y$ on the $dx$ term.

Yes i did that already. What I am after is the answer as to why the method i used above did not work.

 

[Ackbach]

You need to exponentiate the integral of $(M_y-N_x)/N$. That is, in general,
$$u(x)=\exp\left[\int\left(\frac{M_y-N_x}{N}\right)dx\right].$$
I get $u(x)=e^{-x^2}$. See if you agree.

 

[MarkFL]

Solve the ode

$$(y-2x^2y)dx +xdy = 0$$

The equation is in exact form $$Q(x,y)dx+ P(x,y)dy =0$$

When I test for exactness it fails. Then I used the technique $$\frac{M_y-N_x}{N}$$

I get

$u(x)=-2x$ as my integrating factor.

But I end still end up with a non-exact d.e why is that?

$$(-2xy+4x^3y)dx-2x^2dy=0$$

I agree with Ackbach that treating the given ODE as separable is the way to go here. However, if you wish to compute an integrating factor, you should use:

\(\displaystyle \mu(x)=\exp\left(\int -2x\,dx\right)=e^{-x^2}\)

And now, your ODE becomes:

\(\displaystyle \left(e^{-x^2}y\left(1-2x^2\right)\right)dx+\left(xe^{-x^2}\right)dy=0\)

And you can see by inspection this ODE is exact. :)