Differential equation 5y4y' = x2y' + 2xy

[astrofunk21]

Homework Statement

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 I am not sure what to do.

 

[vela]

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)?

 

[astrofunk21]

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)?

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

Correct or no?

 

[vela]

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).

 

[astrofunk21]

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).

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?

 

[astrofunk21]

Could you help with another? The differential equation is:

y' = 2(xy' + y)y³

 

[vela]

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

 

[astrofunk21]

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

F(x,y) = y⁵-x²y + C ?

 

[vela]

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.

 

[astrofunk21]

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.

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

 

[vela]

You could start by checking if it's exact.

 

[astrofunk21]

No it's not, I got

N=1-2xy³
and
M=-2y⁴

which their partials don't equal each other

 

[vela]

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

 

[SammyS]

Could you help with another? The differential equation is:

y' = 2(xy' + y)y³

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.