Solving 1st order DE

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Jul 30th 2013, 05:26 PM
heatly

Solving 1st order DE

How would you go about solving this DE: dy/dx=2xy/(x^2-3y^2) .....could you turn it into an exact differential....ie 2xy dx- (x^2-3y^2) dy=0.....I don't think this will work.
Jul 30th 2013, 08:18 PM
Prove It

Re: Solving 1st order DE

Perhaps a substitution of the form $\displaystyle \begin{align*} v = \frac{y}{x} \implies y = v\,x \implies \frac{dy}{dx} = v + x\,\frac{dv}{dx} \end{align*}$ , giving

$\displaystyle \begin{align*} \frac{dy}{dx} &= \frac{2\,x\,y}{x^2 - 3y^2} \\ v + x\,\frac{dv}{dx} &= \frac{2\,x\,v\,x}{x^2 - 3\left( v\,x \right) ^2} \\ v + x\,\frac{dv}{dx} &= \frac{2\,x^2\,v}{x^2 \left( 1 - 3v^2 \right) } \\ v + x\,\frac{dv}{dx} &= \frac{2\,v}{1 - v^2} \\ x\,\frac{dv}{dx} &= \frac{2\,v }{1 - v^2} - v \\ x\,\frac{dv}{dx} &= \frac{ v + v^3}{1 - v^2} \\ x\,\frac{dv}{dx} &= \frac{v \left( 1 + v^2 \right) }{ 1 - v^2} \\ \frac{1 - v^2}{v\left( 1 + v^2 \right) } \,\frac{dv}{dx} &= \frac{1}{x} \end{align*}$

Since the variables have been separated, you can now solve using integration.
Jul 30th 2013, 10:33 PM
heatly

Re: Solving 1st order DE

Thanks for that
Aug 8th 2013, 05:35 PM
HallsofIvy

Re: Solving 1st order DE

Quote:

Originally Posted by heatly

How would you go about solving this DE: dy/dx=2xy/(x^2-3y^2) .....could you turn it into an exact differential....ie 2xy dx- (x^2-3y^2) dy=0.....I don't think this will work.

The reason is, of course, that this NOT an "exact differential". The derivative of 2xy, with respect to y, is 2x while the derivative of -(x^2- 3y^2), with respect to x, is -2x.

Every first order equation, however, has an "integrating factor". Here, that integrating factor is $\frac{1}{y^2}$ . Multiplying that equation by $\frac{1}{y^2}$ gives $2\frac{x}{y}dx- \left(\frac{x^2}{y^2}- 3\right)dy= 0$ . (Notice that this is now clearly in terms of $\frac{x}{y}$ so Prove It's substitution works.)

The derivative of $\frac{2x}{y}$ , with respect to y, is $-\frac{2x}{y^2}$ and the derivative of $-\left(\frac{x^2}{y^2}- 3\right)$ , with respect to x, is also $-\frac{2x}{y^2}$ .

That means that there exist a function, F(x,y), such that $dF= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy= \frac{2x}{y}dx- \left(\frac{x^2}{y^2}- 3\right)dy= 0$ .

Since $\frac{\partial F}{\partial x}= \frac{2x}{y}$ , $F(x, y)= \frac{x^2}{y}+ g(y)$ for some function, g, of y only. Differentiating that with respect to y, $\frac{\partial F}{\partial y}= -\frac{x^2}{y^2}+ g'= -\frac{x^2}{y^2}+ 3$ so that $g(y)= 3y$ . That is, $F(x,y)= \frac{x^2}{y}+ 3y$ and since "dF= 0" means that F is a constant, the solution to the differential equation is $\frac{x^2}{y}+ 3y= C$ .
Aug 18th 2013, 04:14 AM
heatly

Re: Solving 1st order DE

Thanks a lot