Answer to Question #189908 in Differential Equations for Naina

Question

Dx/y^2+z^2=dy/-xy=dz/-xz

Answer 1

We have to solve the given differential equation,

$\dfrac{dx}{y^2+z^2} = \dfrac{dy}{-xy} = \dfrac{dz}{-xz}$

Firstly, we have to solve last two equations:

then,

$\dfrac{dy}{-xy} = \dfrac{dz}{-xz}$

$\dfrac{dy}{y} = \dfrac{dz}{z}$

$logy = logz+logC$

$y = zC$

Now, we will solve first and last equations:

$\dfrac{dx}{y^2+z^2} = \dfrac{dz}{-xz}$

Putting $y = Cz$

$\dfrac{dx}{z^2C^2+z^2} = -\dfrac{dz}{xz}$

$\dfrac{dx}{z(C^2+1)} = -\dfrac{dz}{x}$

$xdx = -(C^2+1)dz$

Integrating both sides

$\int xdx = - \int (C^2+1)dz$

$\dfrac{x^2}{2} = -(C^2+1)z + C_1$