How to integrate this function?

1. Jan 10, 2014

vanceEE

1. The problem statement, all variables and given/known data
$$xy'' = y' + (y')^{3}$$

2. Relevant equations
$$y'' = p' = \frac{dp}{dx}$$
$$y' = p$$

3. The attempt at a solution
$$xy'' = y' + (y')^{3}$$
$$x*\frac{dp}{dx} = p + p^{3}$$
$$\frac{dx}{x} = \frac{dp}{p+p^3}$$
$$ln x + C = ∫\frac{dp}{p(1+p^2)}$$
$$u = p^{2}$$ $$du = 2p dp$$
$$ln x + C = ∫\frac{1}{p(1+p^2)}*\frac{du}{2p}$$
$$ln x + C = ∫\frac{du}{2p^2(1+p^2)}$$
$$ln x + C = ∫\frac{du}{2u(1+u)}$$
$$ln x + C = \frac{1}{2}∫\frac{du}{u} - \frac{1}{2}∫\frac{du}{1+u}$$
$$ln x + C = \frac{1}{2}ln (\frac{p^2}{p^2+1})$$
$$Dx^2 = \frac{p^2}{p^2+1}$$
$$p^2 = \frac{Dx^2}{1-Dx^2}$$
$$y' = \frac{√(Dx^2)}{√(1-Dx^2)}$$

How do I integrate $$y' = \frac{√(Dx^2)}{√(1-Dx^2)} ?$$

2. Jan 10, 2014

LCKurtz

You don't want a u substitution here. Try partial fractions$$\frac 1 {p(1+p^2)}=\frac A p + \frac {Bp+C}{1+p^2}$$

3. Jan 10, 2014

vanceEE

I used partial fractions after u-substitution.
$$∫\frac{dx}{x} = ∫\frac{du}{2u(1+u)}$$
$$\frac{A}{2u} + \frac{B}{1+u} = \frac{1}{2u(1+u)}$$
$$A(1+u) + B(2u) = 1$$
$$A = 1$$ and $$B = -\frac{1}{2}$$
$$∫\frac{dx}{x} = ∫\frac{du}{2u} - \frac{1}{2}∫\frac{du}{1+u}$$
$$∫\frac{dx}{x} = \frac{1}{2} ∫ \frac{du}{u} - \frac{1}{2} ∫ \frac{du}{1+u}$$
$$ln x + C = \frac{1}{2} ln(p^2) -\frac{1}{2}ln(p^2+1)$$
Therefore, $$Dx^2 = \frac{p^2}{p^2+1}$$
Since $$∫\frac{du}{2u(1+u)} \equiv∫\frac{dp}{p(1+p^2)}$$
$$du = 2p dp → u = p^2$$
$$\frac{du}{2u(1+u)} \equiv \frac{2pdp}{2p^2(1+p^2)} \equiv \frac{dp}{p(1+p^2)}$$

Last edited: Jan 10, 2014
4. Jan 10, 2014

vanceEE

Can someone show me the steps to integrating $$\frac{dy}{dx} = \frac{√(Dx^2)}{√(1-Dx^2)} ?$$
Attempt:
$$\frac{dy}{dx} = \frac{√(Dx^2)}{√(1-Dx^2)}$$
$$dy = \frac{√(Dx^2)}{√(1-Dx^2)} dx$$
$$y = ∫\frac{√(D)x}{√(1-Dx^2)} dx$$
Should I integrate by parts?

Last edited: Jan 10, 2014
5. Jan 10, 2014

vanceEE

Is this correct?
$$\frac{dy}{dx} = \frac{√(Dx^2)}{√(1-Dx^2)}$$
$$dy = \frac{√(Dx^2)}{√(1-Dx^2)} dx$$
$$y = ∫\frac{√(D)x}{√(1-Dx^2)} dx$$
$$y = √(D)∫\frac{x}{√(1-Dx^2)} dx$$
$$u = 1-Dx^2$$
$$du = -2Dx$$
$$y = -\frac{1}{2√(D)}∫\frac{du}{√(u)}$$
$$y = -\frac{1}{2√(D)}* 2√(u) + E$$
$$y = -\frac{1}{2√(D)}* 2√(1-Dx^2) + E$$
$$y = E -\frac{2√(1-Dx^2)}{2√(D)}$$
$$y = E -\frac{√(1-Dx^2)}{√(D)}$$

Last edited: Jan 10, 2014
6. Jan 10, 2014

haruspex

How about x = Bsin(θ) for some suitable B?

7. Jan 10, 2014

Staff: Mentor

The above should be du = -2Dx dx.
Because of the factor of x in the numerator, an ordinary substitution works here. Other than the minor error I pointed out, I don't see anything else wrong, but I didn't check that closely. At this point, you can verify that your work is correct by differentiating to verify that your solution satisifies the differential equation. You don't need us to do that for you.

8. Jan 10, 2014

Staff: Mentor

The above should be du = -2Dx dx.
Because of the factor of x in the numerator, an ordinary substitution works here. Other than the minor error I pointed out, I don't see anything else wrong, but I didn't check that closely. At this point, you can verify that your work is correct by differentiating to verify that your solution satisifies the differential equation. You don't need us to do that for you.

9. Jan 10, 2014

vanceEE

Thank you for pointing this out for me Mark44 :-)