# Differential equation 5y4y' = x2y' + 2xy

1. Feb 19, 2013

### astrofunk21

1. The problem statement, all variables and given/known data
Find the general solution of:

5y4y' = x2y' + 2xy

2. The attempt at a solution

Well I've so far tried to simplify by making the equation really:

(5y4-x2)y' - 2xy = 0

Now this will let us use exact equations such as:

N(x,y)= 5y4-x2
and
M(x,y)= -2xy

since ∂N=∂M

After this though im not sure what to do.

2. Feb 19, 2013

### vela

Staff Emeritus
Re: differential equation

If you have $\Phi(x,y)=0$, when you differentiate it, you get
$$\frac{\partial\Phi}{\partial x}dx + \frac{\partial\Phi}{\partial y}dy = 0.$$ Compare that to what you have
$$(-2xy)\,dx + (5y^4-x^2)\,dy = 0.$$ How would you (partially) recover $\Phi$ from M(x,y) or N(x,y)?

3. Feb 19, 2013

### astrofunk21

Re: differential equation

I then took the integral of M(x,y) thus giving me f-x2y

Correct or no?

4. Feb 19, 2013

### vela

Staff Emeritus
Re: differential equation

That's good, so you have $\Phi=-x^2y + f(y)$. Now differentiate that with respect to y and compare the result to N(x,y).

5. Feb 19, 2013

### astrofunk21

Re: differential equation

finding the solution to the f(y) gives us y5 then plugging this into the integral part we get a solution of:

F(x,y) = y5-x2y

Is that it?

6. Feb 19, 2013

### astrofunk21

Could you help with another? The differential equation is:

y' = 2(xy' + y)y3

7. Feb 19, 2013

### vela

Staff Emeritus
Almost. You just need to include the constant of integration.

8. Feb 19, 2013

### astrofunk21

F(x,y) = y5-x2y + C ?

9. Feb 19, 2013

### vela

Staff Emeritus
Yup, that's right, and F(x,y) should be set to 0. Remember you're trying to find y(x), not F(x,y). F(x,y)=0 specifies y implicitly as a function of x.

10. Feb 19, 2013

### astrofunk21

Thanks! Could you help with the other equation I posted? This one I don't even have a clue how to start

11. Feb 19, 2013

### vela

Staff Emeritus
You could start by checking if it's exact.

12. Feb 19, 2013

### astrofunk21

No it's not, I got

N=1-2xy3
and
M=-2y4

which their partials don't equal eachother

13. Feb 19, 2013

### vela

Staff Emeritus
In that case, try finding an integrating factor that'll make it exact.

14. Feb 19, 2013

### SammyS

Staff Emeritus
It would have been better to start a new thread for this new problem.

(xy' + y) is d(xy)/dx .

See if you can use that.