Answer to Question #289448 in Differential Equations for Pankaj

Question

Using Charpit’s method, solve the equation:

zp²-y²p+y²q =0

Answer 1

$\displaystyle zp^2-y^2p+y^2q=0\\ Let\ f(x,y,z,p,q)=zp^2-y^2p+y^2q,\ then\\ f_x=0,\ f_y=-2yp+2yq,\ f_z=p^2,\ f_p=2zp-y^2,\ f_q=y^2\\ \text{Now, Charpit's auxiliary equations are; }\\ \frac{dx}{-f_p}=\frac{dy}{-f_q}=\frac{dz}{-pf_p-qf_q}=\frac{dp}{f_x+pf_z}=\frac{dq}{f_y+qf_z}\\ \Rightarrow\frac{dx}{y^2-2pz}=\frac{dy}{-y^2}=\frac{dz}{py^2-2p^2z-qy^2}=\frac{dp}{p^3}=\frac{dq}{2yq-2yp+qp^2}\\ \text{From the second and fourth ratio, we get:} \frac{dy}{-y^2}=\frac{dp}{p^3}\Rightarrow p=\sqrt{\frac{y}{ay-2}}, \text{where a is an ar-}\\ \text{bitrary constant.}\\ \Rightarrow0=zp^2-y^2p+y^2q=z\left(\sqrt{\frac{y}{ay-2}}\right)^2-y^2\left(\sqrt{\frac{y}{ay-2}}\right)+y^2q\\ \Rightarrow q=\sqrt{\frac{y}{ay-2}}-\frac{z}{y(ay-2)}\\ But,\ dz=pdx+qdy=\left(\sqrt{\frac{y}{ay-2}}\right) dx+\left(\sqrt{\frac{y}{ay-2}}-\frac{z}{y(ay-2)}\right)dy\\ \text{Multiply both sides by}\ y\sqrt{ay-2}, \text{we have;}\\ (y\sqrt{ay-2})dz=y\sqrt{y}\ dx+y\sqrt{y}\ dy-\frac{z\sqrt{ay-2}}{ay-2}\ dy\\ \Rightarrow (y\sqrt{ay-2})dz=y^{\frac{3}{2}}\ dx+y^{\frac{3}{2}}\ dy-\frac{z\sqrt{ay-2}}{ay-2}\ dy\\ \Rightarrow (y\sqrt{ay-2})dz=y^{\frac{3}{2}}\ dx+y^{\frac{3}{2}}\ dy-\frac{z}{\sqrt{ay-2}}\ dy\\ \text{Integrating both sides yields:}\\ yz\sqrt{ay-2}=xy^{\frac{3}{2}}+\frac{2y^{\frac{3}{2}}}{5}-\frac{2z\sqrt{ay-2}}{a}+b,\ \text{(where b is an arbitrary constant)}\\ \text{which is the solution.}$