How do we solve the following differential equation:

$$\frac{dy}{dx}+ yx^2 = \frac{7x^2}{y}$$

I was hoping to use integrating factors, however that seems inapplicable as the y term is present on both sides of the equation.

Would greatly appreciate any help.

  • $\begingroup$ Do you mean $\frac{dy}{dx}$? $\endgroup$ – André Nicolas Jul 14 '15 at 20:15
  • $\begingroup$ Yes, wasn't sure how to type it out. Thanks for clearing that! $\endgroup$ – abstract Jul 14 '15 at 20:17
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    $\begingroup$ If you mean $\frac{dy}{dx}$, then the equation is separable. $\endgroup$ – André Nicolas Jul 14 '15 at 20:18
  • $\begingroup$ Oh my, should have noticed that... Thank you! $\endgroup$ – abstract Jul 14 '15 at 20:21
  • $\begingroup$ dy/dx + xx(y - 7/y) = 0 => y dy/dx + xx(y*y - 7) = 0 $\endgroup$ – sarker306 Jul 14 '15 at 20:21

The equation is separable. We can rewrite as $$\frac{dy}{dx}=x^2\left(\frac{7}{y}-y\right),$$ and then as $$\frac{y\,dy}{7-y^2}=x^2\,dx.$$ Now integrate.


If you want to use Integrating factor here, you can start with multiplying by $2y$ and rearranging into

$$2y\frac{dy}{dx} + 2x^{2}(y^{2} - 7) = 0$$

Now, let $v = y^{2} - 7$

Then, $$\frac{dv}{dx} = 2y\frac{dy}{dx}$$ Hence, the equation becomes, $$\frac{dv}{dx} + 2x^{2}v = 0$$ Now, it is a Linear Differential Equation, which has defined Integrating Factor.

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    $\begingroup$ A thing to keep in mind: $v$ does not uniquely determine $y$. This isn't actually a problem, because $y$ will not pass through zero on the interval where the solution to any given IVP is defined, but you still have to be careful about it. The same problem happens with separation of variables (the $y$ integration involves a logarithm, and you have to be careful to pick the right logarithm for your initial condition). $\endgroup$ – Ian Jul 14 '15 at 20:31
  • $\begingroup$ Thank you for reminding that. I forget to check for superfluous solutions a lot. Indeed, in this case, $v$ can be shown to be an exponential function, multiplied by a constant C, i.e. $v = Ce^{-\frac{x^{3}}{3}}$ . If we want $y$ to be defined, we should add bounds over C, so that $Ce^{-\frac{x^{3}}{3}} + 7 \geq 0$ $\endgroup$ – sarker306 Jul 14 '15 at 20:39
  • $\begingroup$ Oops, forgot to divide by 2, adding now. Thanks :) $\endgroup$ – sarker306 Jul 14 '15 at 22:38

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