Question

1. Eliminate a, b and c from, z=ax+by+cxy

2. Eliminate a & b from az+b=a²x+y

3. Solve z(x-y)=x²p-y²q

3. Solve 2p+3q=1

Answer 1

(i) $z = ax+by+cxy$

partially differentiating with x and y

$\frac{\partial z}{\partial x} = p = a +cy$ (1)

$\frac{\partial z}{\partial y} = q = b+cx$ (2)

Again differentiating partially first with y, $\frac{\partial^2 z}{\partial x \partial y} = s =c$ (3)

From (1), (2) and (3), $z = (p-sy)x+(q-sx)y+sxy = px+qy-sxy$

(ii) $az+b =a^2x+y$

Partially differentiating with x and y

$\frac{\partial z}{\partial x} = p = a$ , and $a\frac{\partial z}{\partial y} = 1 \implies pq=1$

(iii) $z(x-y) = x^2p-y^2q$

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

Taking first two, $\frac{dx}{x^2}=\frac{dy}{-y^2} \implies c = \frac{1}{x}+\frac{1}{y}$

taking last two terms, $\frac{dy}{-y^2}=\frac{dz}{z(x-y)}$

$\frac{dy}{-y^2}=\frac{dz}{z(\frac{y}{cy-1}-y)} \implies \frac{c}{cy-1}dy = \frac{1}{z}dz$

$z = a(cy-1) \implies a = \frac{xz}{y}$

so solution of the equation is, $f((\frac{1}{x}+\frac{1}{y}),\frac{xz}{y}) = 0$

(iv) $2p+3q =1$

$\frac{dx}{2}=\frac{dy}{3}=\frac{dz}{1}$

integrating first two, $\frac{1}{2}x + c = \frac{1}{3}y \implies c = \frac{1}{3}y-\frac{1}{2}x$

Integrating last two terms, $\frac{1}{3}y + b = z \implies b = z-\frac{1}{3}y$

So solution is, $f((\frac{1}{3}y-\frac{1}{2}x ),(z-\frac{1}{3}y )) = 0$