Answer to Question #189944 in Differential Equations for AMIT YADAV

Question

solve: dx/(y^2 + z^2) = dy/-xy = dz/ -xz

Answer 1

Given differential equation is-

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

Taking Last two terms-

$\dfrac{dy}{-xy}=\dfrac{dz}{-xz}\\[9pt]\Rightarrow \dfrac{dy}{y}=\dfrac{dz}{z}$

Integrate the above equation and we get-

$lny=lnz+lnc_1\Rightarrow lny=lnc_1z\Rightarrow y=c_1z~~~~~-(1)$

Also $c_1=\dfrac{y}{z}~~~~~~-(2)$

Now Taking first and last terms-

$\dfrac{dx}{y^2+z^2}=\dfrac{dz}{-xz}\\[9pt]\Rightarrow \dfrac{ xdx}{z^2c_1^2+z^2}=\dfrac{dz}{-z}~~~~~~~~~~~~~~~~~~~~\text{ from eqn.(1)}\\[9pt]\Rightarrow xdx=-(c_1^2+1)zdz$

Integrating Both the sides-

$\dfrac{x^2}{2}=\dfrac{-z^2}{2}(c_1+1)^2+c_2$

$c_2=\dfrac{x^2}{2}+\dfrac{z^2(\dfrac{y}{z}+1)^2}{2}~~~~~~~~~~\text{from eqn.(2)}\\[9pt]=\dfrac{x^2}{2}+\dfrac{(y+z)^2}{2}~~~~~~~~~-(3)$

The solution of the equation is-

$\phi(c_1,c_2)=0 \\[9pt] \phi(\dfrac{y}{z},\dfrac{(y+z)^2}{2})=0$