Question

Solve z_x+z_y = z² with the initial conditions z(x, 0)=f(x)

Answer 1

P=1

Q=1

R=z²

$\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}$

$\frac{dx}{1}=\frac{dy}{1}=\frac{dz}{z^2}$

Multiplyers: -z²,1,1

-z²dx+dy+dz=0

-z²x+y+z=C₁

Multiplyers: 1,-z²,1

dx-z²dy+dz=0

x-z²y+z=C₂

The general solution is,

$\phi(c_{1},c_{2})=0$

$\phi( -z^2x+y+z),(x-z^2y+z)=0$

-z²x+z+y-C1=0

D=1+4x(y-C₁)

$z_1=\frac{-1+\sqrt{1+4x(y-C_1)}}{-2x}=\frac{-1+\sqrt{1-4xC_1}}{-2x}=f(x)$

$\sqrt{1-4xC_1}=-2xf(x)+1$

$4xC_1=-{(-2xf(x)+1)}^2+1$

$C_1(1)=\frac{-{(-2xf(x)+1)}^2+1}{4x}$

$z_2=\frac{-1-\sqrt{1+4x(y-C_1)}}{-2x}=\frac{-1-\sqrt{1-4xC_1}}{-2x}=f(x)$

$\sqrt{1-4xC_1}=2xf(x)-1$

$4xC_1=-{(2xf(x)-1)}^2+1$

$C_1(2)=\frac{-{(2xf(x)-1)}^2+1}{4x}$

-z²y+z+x-C₂=0

As y=0

z=C₂ - x= f(x)

C₂=f(x) + x