# Homogeneous Differential Equation

#### BesselEquation

1. The problem statement, all variables and given/known data
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...

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#### kuruman

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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

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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

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$$\int \frac{du}{u\sqrt{u^2-u}}=\int \frac{du}{u^2\sqrt{1-1/u}}=\int \frac{d(1-1/u)}{\sqrt{1-1/u}}=\int 2d\sqrt{1-1/u}$$
This is not quite right; it should be
$$\int \frac{d\:\text{anything}}{\sqrt{\text{anything}} }= 2 \sqrt{ \text{anything}}$$

#### Delta2

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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

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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.

"Homogeneous Differential Equation"

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