Answer to Question #105589 in Differential Equations for Khushi

Question

Find the partial differential equation arising from f[z/x^3,y/z]=0, where f is an arbitrary function of the arguments. Also find the general solution of the PDE obtained.

Answer 1

Let $F(x,y,z)=f\left(\frac{z}{x^3},\frac{y}{z}\right)$. Then $F'_x=f'_{x_1}\left(-3\frac{z}{x^4}\right)+f'_{x_2}\cdot 0=-3f'_{x_1}\frac{z}{x^4}$, $F'_y=f'_{x_1}\cdot 0+f'_{x_2}\frac{1}{z}=f'_{x_2}\frac{1}{z}$, $F'_z=f'_{x_1}\frac{1}{x^3}-f'_{x_2}\frac{y}{z^2}$.

Since our function $z(x,y)$ is given by the equation $F(x,y,z)=0$, by the implicit function theorem we obtain $z'_x=-\frac{F'_x}{F'_z}=\frac{3f'_{x_1}\frac{z}{x^4}}{f'_{x_1}\frac{1}{x^3}-f'_{x_2}\frac{y}{z^2}}$ and $z'_y=-\frac{F'_y}{F'_z}=\frac{-f'_{x_2}\frac{1}{z}}{f'_{x_1}\frac{1}{x^3}-f'_{x_2}\frac{y}{z^2}}$.

Rewrite it: $\frac{x}{3z}z'_x=\frac{f'_{x_1}\frac{1}{x^3}}{f'_{x_1}\frac{1}{x^3}-f'_{x_2}\frac{y}{z^2}}$ and $\frac{y}{z}z'_y=\frac{-f'_{x_2}\frac{y}{z^2}}{f'_{x_1}\frac{1}{x^3}-f'_{x_2}\frac{y}{z^2}}$, so $\frac{x}{3z}z'_x+\frac{y}{z}z'_y=1$ or $xz'_x+3yz'_y=3z$.

For obtain a general solution of this equation, we write the following equation: $\frac{dx}{x}=\frac{dy}{3y}=\frac{dz}{3z}$.

1)Solve $\frac{dx}{x}=\frac{dz}{3z}$ . We have $3zdx-xdz=0$. Multiplying it by $\frac{x^2}{z^2}$ , we obtain $\frac{3x^2zdx-x^3dz}{z^2}=0$ or $d\left(\frac{x^3}{z}\right)=0$. So we obtain a first integral $\frac{x^3}{z}=C_1$.

2)Solve $\frac{dy}{3y}=\frac{dz}{3z}$ . We have $ydz-zdy=0$. Multiplying it by $\frac{1}{z^2}$, we obtain $\frac{ydz-zdy}{z^2}=0$ or $-d\left(\frac{y}{z}\right)=0$ . So we obain a second integral $\frac{y}{z}=C_2$.

Therefore the general solution of $xz'_x+3yz'_y=3z$ is $g\left(\frac{x^3}{z},\frac{y}{z}\right)=0$