Question

Solve the differential equation by jacobi's method (p²+q²)y=qz

Answer 1

$(p^2+q^2)y=qz\\ yp^2+yq^2=qz\\$

Let $p=-\frac{u_x}{u_z},q=-\frac{u_y}{u_z}$ .

$y(\frac{u_x^2}{u_z^2})+y(\frac{u_y^2}{u_z^2})=-\frac{u_y}{u_z}.z\\ yu_x^2+yu_y^2=-zu_yu_z\\ \implies f(x,y,z,u_x,u_y,u_z)=yu_x^2+yu_y^2+zu_yu_z=0$

$f_{u_x}=2yu_x,f_{u_y}=2yu_y+zu_z,f_{u_z}=zu_y,f_x=0,f_y=u_x^2+u_y^2,f_z=u_yu_z$ .

The auxilliary equation is given as;

$\frac{dx}{f_{u_x}}=\frac{dy}{f_{u_y}}=\frac{dz}{f_{u_z}}=\frac{du_x}{-f_x}=\frac{du_y}{-f_y}=\frac{du_z}{-f_z}\\ \frac{dx}{2yu_x}=\frac{dy}{2yu_y+zu_z}=\frac{dz}{zu_y}=\frac{du_x}{0}=\frac{du_y}{-(u_x^2+u_y^2)}=\frac{du_z}{-u_yu_z}$

From the fraction above, we see that

$du_x=0 \implies u_x=a$ \\

Also,

$\frac{dz}{zu_y}=\frac{du_z}{-u_yu_z}\\ \frac{dz}{z}+\frac{du_z}{u_z}=0\\ \text{Integrate}\\ \ln z +\ln u_z=\ln b\\ u_z=\frac{b}{z}$

Substitute the values for $u_x$ and $u_z$ into $f$

$a^2y+yu_y^2+bu_y=0\\ yu_y^2+bu_y+a^2y=0\\ u_y=\frac{-b \pm \sqrt{b^2-4a^2y^2}}{2y}$

Take the positive value\\

$u_y=\frac{-b+ \sqrt{b^2-4a^2y^2}}{2y}$

$du=u_xdx+u_ydy+u_zdz\\ du=adx+\frac{-b+ \sqrt{b^2-4a^2y^2}}{2y}dy+\frac{b}{z}dz\\ \text{integrate}\\ u=ax+\frac{1}{2}\left(\sqrt{b^2-4a^2y^2}-b\tanh^{-1}\left(\frac{\sqrt{b^2-4a^2y^2}}{b}\right)\right)+b \ln z+c$

Let u=c

$ax+\frac{1}{2}\left(\sqrt{b^2-4a^2y^2}-b\tanh^{-1}\left(\frac{\sqrt{b^2-4a^2y^2}}{b}\right)\right)+b\ln z=0$