Homogeneous Differential Equation

[BesselEquation]

Homework Statement

Solve the following differential equation:
y' = y / [ x + √(y^2 - xy)]

2. The attempt at a solution
Using the standard method for solving homogeneous equations, setting u = y/x, I arrive at the following:

± dx/x = [1±√(u^2-u) ]/ [u√(u^2-u)] which in turn, I get the following integral after simplifying:

∫du/[u√(u^2-u)], seems quite unsolvable to me...

 

[Delta2]

Well , according to wolfram the integral $$\int \frac{du}{u\sqrt{u^2-u}}=2\sqrt{\frac{u-1}{u}}$$

https://www.wolframalpha.com/input/?i=integral+du/(usqrt(u^2-u))

Don't have any ideas right now on how you actually calculate this integral...

 

[kuruman]

Note that$$\frac{1}{u\sqrt{u^2-u}}=\frac{1}{u^{3/2}(u-1)^{1/2}}$$Here's where the guessing and experience with derivatives comes in. The idea is to write this as a perfect differential which would make the integration trivial. You know that the term $u^{3/2}$ in the denominator arises by taking the derivative of $1/u^{1/2}$ and the term $(u-1)^{1/2}$ in the denominator arises by taking the derivative of $(u-1)^{1/2}$. You also know the product rule of differentiation. What do you get when you take the derivative of $\frac{(u-1)^{1/2}}{u^{1/2}}~$?

 

[lurflurf]

Well , according to wolfram the integral $$\int \frac{du}{u\sqrt{u^2-u}}=2\sqrt{\frac{u-1}{u}}$$

$$\int \frac{du}{u\sqrt{u^2-u}}=\int \frac{du}{u^2\sqrt{1-1/u}}=\int \frac{d(1-1/u)}{\sqrt{1-1/u}}=\int 2d\sqrt{1-1/u}$$

 

[BesselEquation]

Woah alright, great...

 

[Ray Vickson]

$$\int \frac{du}{u\sqrt{u^2-u}}=\int \frac{du}{u^2\sqrt{1-1/u}}=\int \frac{d(1-1/u)}{\sqrt{1-1/u}}=\int 2d\sqrt{1-1/u}$$

This is not quite right; it should be
$$\int \frac{d\:\text{anything}}{\sqrt{\text{anything}} }= 2 \sqrt{ \text{anything}}$$

 

[Delta2]

This is not quite right; it should be
$$\int \frac{d\:\text{anything}}{\sqrt{\text{anything}} }= 2 \sqrt{ \text{anything}}$$

I don't understand where you disagree, isn't it $\frac{d f}{\sqrt{f}}=2d(\sqrt{f)}$ and then the integral operator cancels out with the differential operator d and leaves just $2\sqrt{f}$

 

[Ray Vickson]

I don't understand where you disagree, isn't it $\frac{d f}{\sqrt{f}}=2d(\sqrt{f)}$ and then the integral operator cancels out with the differential operator d and leaves just $2\sqrt{f}$

OK, you are right. Basically, I did not notice that you just left out the last step.