How to integrate this function?

[vanceEE]

Homework Statement

$$xy'' = y' + (y')^{3}$$

Homework Equations

$$ y'' = p' = \frac{dp}{dx}$$
$$ y' = p$$

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

 

[LCKurtz]

Homework Statement

$$xy'' = y' + (y')^{3}$$

Homework Equations

$$ y'' = p' = \frac{dp}{dx}$$
$$ y' = p$$

The Attempt at a Solution

$$xy'' = y' + (y')^{3}$$
$$ x*\frac{dp}{dx} = p + p^{3} $$
$$ \frac{dx}{x} = \frac{dp}{p+p^3} $$
$$ ln x + C = ∫\frac{dp}{p(1+p^2)} $$
$$ u = p^{2} $$ $$du = 2p dp$$

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

 

[vanceEE]

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

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 $$ ∫\frac{du}{2u(1+u)} \equiv∫\frac{dp}{p(1+p^2)}$$
$$ du = 2p dp → u = p^2 $$
$$\frac{du}{2u(1+u)} \equiv \frac{2pdp}{2p^2(1+p^2)} \equiv \frac{dp}{p(1+p^2)} $$

 

[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?

 

[vanceEE]

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

 

[haruspex]

Can someone show me the steps to integrating $$ \frac{dy}{dx} = \frac{√(Dx^2)}{√(1-Dx^2)} ?$$

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

 

[Mark44]

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

The above should be du = -2Dx dx.

$$ y = -\frac{1}{2√(D)}∫\frac{du}{√(u)} $$
$$ y = -\frac{1}{2√(D)}* 2√(u) + E $$
$$ y = -\frac{1}{2√(D)}* 2√(1-Dx^2) + E $$
$$ y = E -\frac{2√(1-Dx^2)}{2√(D)} $$
$$ y = E -\frac{√(1-Dx^2)}{√(D)} $$

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.

 

[Mark44]

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

The above should be du = -2Dx dx.

$$ y = -\frac{1}{2√(D)}∫\frac{du}{√(u)} $$
$$ y = -\frac{1}{2√(D)}* 2√(u) + E $$
$$ y = -\frac{1}{2√(D)}* 2√(1-Dx^2) + E $$
$$ y = E -\frac{2√(1-Dx^2)}{2√(D)} $$
$$ y = E -\frac{√(1-Dx^2)}{√(D)} $$

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.

 

[vanceEE]

The above should be du = -2Dx dx

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