Question

solve the Lagrange partial differential equation x(z+2a)p + (xz+2yz+2ay)q=z(z+a)​

Answer 1

solve the Lagrange partial differential equation x(z+2a)p + (xz+2yz+2ay)q=z(z+a)

Answer 2

Answer:

The Lagrange partial differential equation $x(z+2a)p + (xz+2yz+2ay)q=z(z+a)$is given by ​$\frac{y}{z}-\frac{x}{z^{2}}=-\frac{x(z+1)}{z^{2}} \ln \left(\frac{4 x}{z^{2}}\right)-\frac{1}{4}$

Step-by-step explanation:

Ist order linear partial differential equation in its standard form

$Pp + Qq = R---(1)$

where P, Q, R are functions of x, y, z is called Lagrange’s Linear Equation. This equation is

obtained by eliminating arbitrary function f from

$f(u, v) = 0----(2)$

where u, v are functions of x, y, z.

$&\frac{d x}{x(z+2)}=\frac{d y}{x z+2 y z+2 y}=\frac{d z}{z(z+1)}\\&\frac{d x}{x(z+2)}=\frac{d z}{z(z+1)}\\&\frac{d x}{x}=\frac{(z+2) d z}{z(z+1)}\\&\int \frac{d x}{x}=\int \frac{(z+2) d z}{z(z+1)}\\&\int \frac{(z+2) d z}{z(z+1)}=\int \frac{d z}{z+1}+2 \int \frac{d z}{z(z+1)}$

$&\frac{1}{z(z+1)}=\frac{A}{z}+\frac{B}{z+1}\\&A(z+1)+B z=1\\&A+B=0, A=1, B=-1\\&\frac{1}{z(z+1)}=\frac{1}{z}-\frac{1}{z+1}\\&\ln x=\ln (z+1)+2 \ln z-2 \ln (z+1)+\ln c_{1}\\&\ln x=\ln \left(\frac{c_{1} z^{2}}{z+1}\right)\\&\frac{x(z+1)}{z^{2}}=c_{1}$

$&\frac{d y}{x z+2 y z+2 y}=\frac{d z}{z(z+1)}\\&\frac{(z+1) d y}{c_{1} z^{3}+2 y(z+1)^{2}}=\frac{d z}{z(z+1)}\\&\frac{d y}{d z}-\frac{y}{z}=\frac{c_{1} z^{2}}{(z+1)^{2}}\\&y=u v, y^{\prime}=u^{\prime} v+u v^{\prime}\\&u^{\prime} v+u\left(v^{\prime}-\frac{v}{z}\right)=\frac{c_{1} z^{2}}{(z+1)^{2}}\\&v^{\prime}-\frac{v}{z}=0 \Longrightarrow v=z\\&u^{\prime} v=u^{\prime} z=\frac{c_{1} z^{2}}{(z+1)^{2}}$

$&\frac{d u}{d z}=\frac{c_{1} z}{(z+1)^{2}}\\&u=c_{1} \int \frac{z d z}{(z+1)^{2}}\\&\frac{z}{(z+1)^{2}}=\frac{A}{z+1}+\frac{B}{(z+1)^{2}}\\&A(z+1)+B=z\\&A=1, A+B=0, B=-1\\&\frac{z}{(z+1)^{2}}=\frac{1}{z+1}-\frac{1}{(z+1)^{2}}$

$\begin{aligned}&u=c_{1} \int\left(\frac{1}{z+1}-\frac{1}{(z+1)^{2}}\right) d z=c_{1}\left(\ln (z+1)+\frac{1}{z+1}\right)+c_{2} \\&y=u v=c_{1} z\left(\ln (z+1)+\frac{1}{z+1}\right)+c_{2} z \\&y=\frac{x(z+1)}{z}\left(\ln (z+1)+\frac{1}{z+1}\right)+c_{2} z\end{aligned}$

$\frac{y}{z}-\frac{x(z+1)}{z^{2}}\left(\ln (z+1)+\frac{1}{z+1}\right)=c_{2}$

For the curve $\left(s^{2}, 0,2 s\right)$:

$\begin{aligned}&c_{1}=\frac{s^{2}(2 s+1)}{4 s^{2}}=\frac{2 s+1}{4} \\&c_{2}=-\frac{2 s+1}{4}\left(\ln (2 s+1)+\frac{1}{2 s+1}\right)\end{aligned}$

$c_{2}=-c_{1}\left(\ln \left(4 c_{1}\right)+\frac{1}{4 c_{1}}\right)=-c_{1} \ln \left(4 c_{1}\right)-\frac{1}{4}$

Substituting $c_{1} and c_{2}$ we get the integral surface:

$\frac{y}{z}-\frac{x(z+1)}{z^{2}}\left(\ln (z+1)+\frac{1}{z+1}\right)=-\frac{x(z+1)}{z^{2}} \ln \left(\frac{4 x(z+1)}{z^{2}}\right)-\frac{1}{4}$

$\frac{y}{z}-\frac{x}{z^{2}}=-\frac{x(z+1)}{z^{2}} \ln \left(\frac{4 x}{z^{2}}\right)-\frac{1}{4}$