I have
$$\left(2x+ 1-\frac{y^2}{x^2}\right)dx+ \frac{2y}{x}dy= 0$$
which is not exact. However, it could be made exact by using the formula for integrating factors. I have two ways,
With $$M=\left(2x+ 1-\frac{y^2}{x^2}\right)$$ and $$N=\frac{2y}{x}$$ we can form the following integrating factors:
$$\phi(x)=\frac{N_x-M_y}{M}=-\frac{\frac{4y}{x^2}}{2x+1-\frac{y^2}{x^2}}$$
or
$$\psi(x)=\frac{M_y-N_x}{N}=\frac{ (x^3 – x^2 + y^2)}{yx^2}$$
Then multiplying these in, should give an exact form of the ODE. But neither of the two make the ODE exact. Are there other formulas one can use?
Thanks
Best Answer
$$\left(2x+ 1-\frac{y^2}{x^2}\right)dx+ \frac{2y}{x}dy= 0$$ $$\left(2x+1\right)dx+\dfrac {(-y^2dx +{2xy}dy)}{x^2}= 0$$ $$\left(2x+1\right)dx+\dfrac {(-y^2dx +{x}dy^2)}{x^2}= 0$$ $$\left(2x+1\right)dx+d \left (\dfrac {y^2}{x}\right)= 0$$ Integrate.