Differential equation parametrisation with integrating factor

[Perrault]

Homework Statement

Use parametrisation first, derive the equation including y and p = [itex]\frac{dy}{dx}[/itex] and use the integrating factor method to reduce it to an exact equation. Leave the solution in implicit parametric form.

[itex](y')^{3}[/itex] + y[itex]^{2}[/itex] = xyy'

The Attempt at a Solution

I'm really lost at this. I tried writing p=y'
p[itex]^{3}[/itex] + y[itex]^{2}[/itex]=xyp
[itex]\frac{p^{3}+y^{2}}{yp}[/itex] = x
[itex]\frac{p^{3}}{yp}[/itex] + [itex]\frac{y^{2}}{yp}[/itex] = x
[itex]\frac{p^{2}}{y}[/itex] + [itex]\frac{y}{p}[/itex] = x

And I don't really know what to do from there. Some facebook rumors propose that the integrating factor be [itex]\frac{1}{y^{3}}[/itex]

 

[mighty2000]

.

Hello!

Thank you for your post. It seems like you are on the right track with your attempt at a solution. Let me walk you through the steps to solve this problem using the integrating factor method.

Step 1: Parametrize the equation

To start, let's use the parametrization method as suggested in the forum post. This means that we will rewrite the equation in terms of p (the derivative of y with respect to x) and y.

We can rewrite the equation as:

p^3 + y^2 = xy'

Step 2: Rewrite the equation in terms of p and y

Next, we can rewrite the equation as:

p^3 + y^2 - xy' = 0

Step 3: Multiply by the integrating factor

Now, we can multiply both sides of the equation by the integrating factor, which in this case is 1/y^3. This will give us:

(p^3 + y^2 - xy')\frac{1}{y^3} = 0

Step 4: Simplify

Simplifying the left side of the equation, we get:

\frac{p^3}{y^2} + \frac{1}{y} - \frac{x}{y^3} = 0

Step 5: Use the chain rule to rewrite the equation

Next, we can use the chain rule to rewrite the equation as:

\frac{d}{dx}(\frac{p^3}{y^2}) + \frac{d}{dx}(\frac{1}{y}) - \frac{d}{dx}(\frac{x}{y^3}) = 0

Step 6: Apply the product rule

Applying the product rule, we get:

\frac{3p^2p'}{y^2} - \frac{2p^3y'}{y^3} - \frac{y'}{y^2} + \frac{3xy'}{y^4} = 0

Step 7: Simplify

Simplifying further, we get:

\frac{3p^2p'}{y^2} - \frac{2p^3y'}{y^3} - \frac{y'}{y^2} + \frac{3xy'}{y^4} = 0