Suppose that $(y^2+2xy) dx$-$x^2$$dy$=0 has an integrating factor that is a function of $y$ alone $[i.e., μ = μ(y)]$. Find the integrating factor and use it to solve the differential equation.

For this one, I get the integrating factor $\mu$=$\frac{1}{\sqrt y}$, which is correct.

Then I multiply the integrating factor with the original equation, $\frac{2y}{\sqrt y}$$dx$+ $\frac{x+y}{\sqrt y}$$dy$=$0$

Next I try integrating both sides since the ODE is separable, get $\int \frac{2y}{\sqrt y}$$dx$+$\int \frac{x+y}{\sqrt y}$$dy$=$0$

And the result is$\frac{2y}{\sqrt y}$$*x$+$x*(\sqrt y)$+$\frac{2}{3}$*$y^{\frac{3}{2}}$=$F(x,y)$=$C$. But doesn't match the answer.

The correct answer is $\frac{2y}{\sqrt y}$$*x$+$\frac{2}{3}$*$y^{\frac{3}{2}}$=$F(x,y)$=$C$. What am I doing wrong?

  • 1
    $\begingroup$ It's Bernouilli's equation are you sure about the solution of the book ? $\endgroup$ – Aryadeva Jan 20 at 17:46
  • $\begingroup$ oops, I realize that my answer is right, just need to simplify a little. I looked up the wrong question and found the wrong answer. Sorry guys, problem solved. $\endgroup$ – Beacon Jan 20 at 17:52
  • $\begingroup$ no problem Beacon $\endgroup$ – Aryadeva Jan 20 at 17:57

This is Bernouilli's equation. $$(y^2+2xy) dx-x^2dy=0$$ $$y^2dx+y dx^2-x^2dy=0$$ Integrating factor $\mu (y )=\dfrac 1 {y^2}$ $$dx+\dfrac {y dx^2-x^2dy}{y^2}=0$$ $$dx+d\left (\dfrac {x^2}{y} \right)=0$$ $$x+ \left( \dfrac {x^2}{y} \right)=C$$ Are you sure it's the right equation and solution ?

| cite | improve this answer | |
  • $\begingroup$ The solution is $2x \sqrt y$+$\frac{2}{3}$$y^{\frac{3}{2}}$=$C$ $\endgroup$ – Beacon Jan 20 at 17:58
  • 1
    $\begingroup$ For which equation ? The one you posted has not that solution. @Beacon $\endgroup$ – Aryadeva Jan 20 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.