Question

form a PDE by elimination f and g from z= f(x2-y)+g(x2+y)

Answer 1

form a PDE by elimination $f$ and $g$ from $z = f(x^2 - y) + g(x^2 + y)$

Solution

Given $= f(x^{2} - y) + g(x^{2} + y)$ (1)

Differentiating (1) partially with respect to $x$,

$$\frac {\partial z}{\partial x} = f ^ {\prime} (u) * 2 x + g ^ {\prime} (v) * 2 x = 2 x \left(f ^ {\prime} (u) + g ^ {\prime} (v)\right)$$

Where $u = x^2 - y$ and $v = x^2 + y$

Differentiating (1) partially with respect to $y$,

$$\frac {\partial z}{\partial y} = f ^ {\prime} (u) * (- 1) + g ^ {\prime} (v) * 1 = g ^ {\prime} (v) - f ^ {\prime} (u)$$

Differentiating (2) partially with respect to $x$,

$$\frac {\partial^ {2} z}{\partial x ^ {2}} = \left(2 x \left(f ^ {\prime} (u) + g ^ {\prime} (v)\right)\right) ^ {\prime} = 2 \left(f ^ {\prime} (u) + g ^ {\prime} (v)\right) + (2 x) ^ {2} \left(f ^ {\prime \prime} (u) + g ^ {\prime \prime} (v)\right)$$

Differentiating (3) partially with respect to $y$,

$$\frac {\partial^ {2} z}{\partial y ^ {2}} = \left(g ^ {\prime} (v) - f ^ {\prime} (u)\right) ^ {\prime} = \left(f ^ {\prime \prime} (u) + g ^ {\prime \prime} (v)\right)$$

From (2), (4) and (5) we have

$$\frac {\partial^ {2} z}{\partial x ^ {2}} = (2 x) ^ {2} \frac {\partial^ {2} z}{\partial y ^ {2}} + \frac {1}{x} \frac {\partial z}{\partial x}$$

Answer: $\frac{\partial^2z}{\partial x^2} = (2x)^2\frac{\partial^2z}{\partial y^2} +\frac{1}{x}\frac{\partial z}{\partial x}.$