Answer to Question #114851 in Differential Equations for Sheela John

Question

Solve the following
z(p^2+q^2)+px+qy=0

Answer 1

The given differential equation is

$f(x,y,z,p,q)\equiv z(p^2+q^2)+px+qy=0$ ...........(1)

Which is a Charpit's equation.

Therefore , The Charpit's Auxiliary equation are

$\frac{dp}{f_x+pf_z}=\frac{dq}{f_y+qf_z}=\frac{dz}{-pf_p-qf_q}=\frac{dx}{-f_p}=\frac{dy}{-f_q}$

$\implies \frac{dp}{p+p^3+pq^2}=\frac{dq}{q+q^3+qp^2}=\frac{dz}{-p(x+2zp)-q(y+2zq)}=\frac{dx}{-x-2zp}=\frac{dy}{-y-2zq}$

Taking first two fraction ,We get

$$\frac{dp}{p(1+p^2+q^2)}=\frac{dq}{q(1+p^2+q^2)}$$

$\implies \frac{dp}{p}=\frac{dq}{q} \ \implies log(p)=log(q)+log(a)$

Where $log(a)$ is a integration constant.

$\implies \frac{p}{q}=a$ $\implies p=aq$ ........(2)

Putting the value of (2) in equation (1) ,we get

$z(a^2q^2+q^2)+aqx+qy=0$

$\implies z(a^2q+q)+ax+y=0 \implies q=\frac{-(ax+y)}{z(a^2+1)}$

and $p=aq=-a×\frac{ax+y}{z(a^2+1)}$

Now putting the value of p and q in $dz=pdx+qdy$

$\implies (a^2+1)zdz=-a^2xdx-aydx-axdy-ydy$

$\implies (a^2+1)zdz=-a^2xdx-ad(xy)-ydy$

On integrating ,we get

$\frac{(a^2+1)}{2}×z^2 =-\frac{a^2}{2}x^2-axy-\frac{y^2}{2}+\frac{k}{2}$

Where k is a integration constant.

$(a^2+1)×z^2=-a^2x^2-axy-y^2+k$

Which is a complete integral.