# Verify whether y^2 is an IF of dx/y+2x/y^2.dy=0?

Aug 11, 2018

It is not

#### Explanation:

If this is already exact , then:

${\underbrace{\frac{1}{y}}}_{= {f}_{x}} \mathrm{dx} + {\underbrace{\frac{2 x}{y} ^ 2}}_{= {f}_{y}} \mathrm{dy} = {\underbrace{0}}_{\because \mathrm{df} = {f}_{x} \mathrm{dx} + {f}_{y} \mathrm{dy} = 0}$

The mixed partials are not equal, so it is not exact.

• $\left\{\begin{matrix}{f}_{x y} = - \frac{1}{y} ^ 2 \\ {f}_{y x} = \frac{2}{y} ^ 2 \\ {f}_{x y} \ne {f}_{y x}\end{matrix}\right.$

With your suggested I.F., it becomes:

${\underbrace{y}}_{= {f}_{x}} \mathrm{dx} + {\underbrace{2 x}}_{= {f}_{y}} \mathrm{dy} = 0$

• $\left\{\begin{matrix}{f}_{x y} = 1 \\ {f}_{y x} = 2 \\ {f}_{x y} \ne {f}_{y x}\end{matrix}\right.$

Ditto: not exact.

Try instead an I.F. that is ${y}^{3}$:

${\underbrace{{y}^{2}}}_{= {f}_{x}} \mathrm{dx} + {\underbrace{2 x y}}_{= {f}_{y}} \mathrm{dy} = 0$

• $\left\{\begin{matrix}{f}_{x y} = 2 y \\ {f}_{y x} = 2 y \\ \boldsymbol{{f}_{x y} = {f}_{y x}}\end{matrix}\right.$

This is exact.

Integrating:

• $\int \setminus {y}^{2} \setminus \partial x = x {y}^{2} + \alpha \left(y\right)$

• $\int \setminus 2 x y \setminus \partial y = x {y}^{2} + \beta \left(x\right)$

So there is solution:

• $q \quad f \left(x , y\right) = x {y}^{2} = C$