# Error check: Suppose that $(y^2+2xy) dx$-$x^2 dy$=0 has an integrating factor that is a function of $y$ alone $[i.e., μ = μ(y)]$.

Suppose that $$(y^2+2xy) dx$$-$$x^2dy$$=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?

• It's Bernouilli's equation are you sure about the solution of the book ? – Aryadeva Jan 20 at 17:46
• 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. – Beacon Jan 20 at 17:52
• no problem Beacon – 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 ?
• The solution is $2x \sqrt y$+$\frac{2}{3}$$y^{\frac{3}{2}}$=$C$ – Beacon Jan 20 at 17:58