Question

solve by lagrange's method z(p-q) =z^2+(x+y)^2

Answer 1

Standard pde is Pp+Qq=R,

where p=$\frac{ \partial z}{\partial x}$ and q=$\frac{ \partial z}{\partial y}$ .

Comparing the given pde with the standard pde.

We get, P=z, Q=-z, R=z²+(x+y)²

By lagrange method,

$\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}\\ \frac{dx}{z}=\frac{dy}{-z}=\frac{dz}{z^2+(x+y)^2}\\$

By taking first equation,

$\frac{dx}{z}=\frac{dy}{-z}\\ dx=-dy\\ x=-y+c_1\\ x+y=c_1$

By taking second equation,

$\frac{dy}{z}=\frac{dz}{z^2+(x+y)^2}\\ dy=\frac{z}{z^2+c_1^2}dz [ \space \text{from first equation}]\\ dy=\frac{1}{2}\frac{2z}{z^2+c_1^2}dz$

Integrating both side, we get

$y=\frac{1}{2}ln(z^2+c_1^2)+c_2\\ y-ln\sqrt{(z^2+(x+y)^2)}=c_2$

Therefore, the solution is given by

$F(x+y, y-ln\sqrt{(z^2+(x+y)^2)})=0$