Question

dx/(z^2+2y)=dy/(z^2+2x)=dz/-z

Answer 1

Solution

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

From first two

$(z^2+2x)dx=(z^2+2y)dy$

Here z is acting as constant , further integrating both sides we get

$z^2x+x^2+c_1=z^2y+y^2+c_2$ ......(1)

Now taking first and last

$\frac{dx}{ (z^2+2y)}=\frac{dz}{ (-z)}$

Integrating both sides, here y and z will act as constant

We get

$\frac{x}{z^2+2y}+c_3=-ln(z)+c_4$ ..........(2)

Equation (1) and (2) are answers .