Question

Find the complete integral of the equation
∂z/∂x[1+(∂z/∂y) ^2]=∂z/∂y(z-a)

Answer 1

Given differential equation $\frac{∂z}{∂x}[1+(\frac{∂z}{∂y}) ^2]=\frac{∂z}{∂y}(z-a)$

we know that, $p = \frac{∂z}{∂x}$ and $q = \frac{∂z}{∂y}$

Let $u = x + by$

Then $p = \frac{dz}{du}$ and $q = b\frac{dz}{du}$

Then equation will be

$\frac{dz}{du}[1 +b^2(\frac{dz}{du})^2 ] = b(\frac{dz}{du})(z-a)$

$\frac{dz}{du} = \frac{\sqrt{(bz-ab-1)}}{b}$

Integrating equation,

$\int\frac{bdz}{\sqrt(bz-ab-1)} = \int du$

$2\sqrt{(bz-ab-1)} = u + C$

then

$4(bz-ab-1) = (u+C)^2$

putting value of u,

$4(bz-ab-1) = (x+by+C)^2$

which is the required solution.