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.
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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.
Perhaps a substitution of the form $\displaystyle \displaystyle \begin{align*} v = \frac{y}{x} \implies y = v\,x \implies \frac{dy}{dx} = v + x\,\frac{dv}{dx} \end{align*}$, giving
$\displaystyle \displaystyle \begin{align*} \frac{dy}{dx} &= \frac{2\,x\,y}{x^2 - 3y^2} \\ v + x\,\frac{dv}{dx} &= \frac{2\,x\,v\,x}{x^2 - 3\left( v\,x \right) ^2} \\ v + x\,\frac{dv}{dx} &= \frac{2\,x^2\,v}{x^2 \left( 1 - 3v^2 \right) } \\ v + x\,\frac{dv}{dx} &= \frac{2\,v}{1 - v^2} \\ x\,\frac{dv}{dx} &= \frac{2\,v }{1 - v^2} - v \\ x\,\frac{dv}{dx} &= \frac{ v + v^3}{1 - v^2} \\ x\,\frac{dv}{dx} &= \frac{v \left( 1 + v^2 \right) }{ 1 - v^2} \\ \frac{1 - v^2}{v\left( 1 + v^2 \right) } \,\frac{dv}{dx} &= \frac{1}{x} \end{align*}$
Since the variables have been separated, you can now solve using integration.
Thanks for that
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.Quote:
Originally Posted by heatly
Every first order equation, however, has an "integrating factor". Here, that integrating factor is $\displaystyle \frac{1}{y^2}$. Multiplying that equation by $\displaystyle \frac{1}{y^2}$ gives $\displaystyle 2\frac{x}{y}dx- \left(\frac{x^2}{y^2}- 3\right)dy= 0$. (Notice that this is now clearly in terms of $\displaystyle \frac{x}{y}$ so Prove It's substitution works.)
The derivative of $\displaystyle \frac{2x}{y}$, with respect to y, is $\displaystyle -\frac{2x}{y^2}$ and the derivative of $\displaystyle -\left(\frac{x^2}{y^2}- 3\right)$, with respect to x, is also $\displaystyle -\frac{2x}{y^2}$.
That means that there exist a function, F(x,y), such that $\displaystyle dF= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy= \frac{2x}{y}dx- \left(\frac{x^2}{y^2}- 3\right)dy= 0$.
Since $\displaystyle \frac{\partial F}{\partial x}= \frac{2x}{y}$, $\displaystyle F(x, y)= \frac{x^2}{y}+ g(y)$ for some function, g, of y only. Differentiating that with respect to y, $\displaystyle \frac{\partial F}{\partial y}= -\frac{x^2}{y^2}+ g'= -\frac{x^2}{y^2}+ 3$ so that $\displaystyle g(y)= 3y$. That is, $\displaystyle F(x,y)= \frac{x^2}{y}+ 3y$ and since "dF= 0" means that F is a constant, the solution to the differential equation is $\displaystyle \frac{x^2}{y}+ 3y= C$.
Thanks a lot
