1. **Problem 1: Eliminate arbitrary constants a and b** (a) Given $$z = ax + by + a^2 b^2$$ We want to eliminate constants $a$ and $b$ to find a PDE. (b) Given $$z = a x^2 + b y^2$$ Similarly, eliminate $a$ and $b$. --- 2. **Problem 2: Eliminate arbitrary function to find first order PDE** (a) Given $$z = f(x^2 + y^2)$$ (b) Given $$z = x f\left(\frac{y}{x}\right)$$ --- 3. **Problem 3: Find general solution of PDE** $$(y + z)p + (z + x)q = x + y$$ where $p = \frac{\partial z}{\partial x}$ and $q = \frac{\partial z}{\partial y}$. --- ### Step-by-step solutions: **1(a).** - Given $$z = a x + b y + a^2 b^2$$ - Differentiate partially w.r.t. $x$: $$z_x = a$$ - Differentiate partially w.r.t. $y$: $$z_y = b$$ - From these, $a = z_x$ and $b = z_y$. - Substitute back into original equation: $$z = z_x x + z_y y + (z_x)^2 (z_y)^2$$ - Rearranged, this is the PDE governing $z$: $$z - x z_x - y z_y = (z_x)^2 (z_y)^2$$ **1(b).** - Given $$z = a x^2 + b y^2$$ - Differentiate w.r.t. $x$: $$z_x = 2 a x$$ - Differentiate w.r.t. $y$: $$z_y = 2 b y$$ - Solve for $a$ and $b$: $$a = \frac{z_x}{2 x}, \quad b = \frac{z_y}{2 y}$$ - Substitute into original: $$z = x^2 \frac{z_x}{2 x} + y^2 \frac{z_y}{2 y} = \frac{x z_x}{2} + \frac{y z_y}{2}$$ - Multiply both sides by 2: $$2 z = x z_x + y z_y$$ - This is the PDE: $$2 z = x z_x + y z_y$$ **2(a).** - Given $$z = f(x^2 + y^2)$$ - Let $$u = x^2 + y^2$$, so $$z = f(u)$$ - Compute derivatives: $$z_x = f'(u) \cdot 2x$$ $$z_y = f'(u) \cdot 2y$$ - From these, $$y z_x - x z_y = y (2x f'(u)) - x (2y f'(u)) = 2 f'(u)(x y - x y) = 0$$ - Hence the PDE is: $$y z_x - x z_y = 0$$ **2(b).** - Given $$z = x f\left(\frac{y}{x}\right)$$ - Let $$v = \frac{y}{x}$$, so $$z = x f(v)$$ - Compute derivatives: $$z_x = f(v) + x f'(v) \cdot \frac{\partial v}{\partial x}$$ $$\frac{\partial v}{\partial x} = \frac{-y}{x^2} = -\frac{v}{x}$$ So, $$z_x = f(v) - v f'(v)$$ - Compute $z_y$: $$z_y = x f'(v) \cdot \frac{\partial v}{\partial y}$$ $$\frac{\partial v}{\partial y} = \frac{1}{x}$$ So, $$z_y = f'(v)$$ - Form PDE: $$x z_y + y z_x = x f'(v) + y (f(v) - v f'(v)) = x f'(v) + y f(v) - y v f'(v)$$ - Since $v = y/x$, $y v = y \cdot \frac{y}{x} = \frac{y^2}{x}$ - Simplify: $$x z_y + y z_x = x f'(v) + y f(v) - \frac{y^2}{x} f'(v) = y f(v) + f'(v) \left(x - \frac{y^2}{x}\right)$$ - Note that $$x - \frac{y^2}{x} = \frac{x^2 - y^2}{x}$$ - The PDE is: $$x z_y + y z_x = y f(v) + f'(v) \frac{x^2 - y^2}{x}$$ - Since $f$ is arbitrary, the PDE that eliminates $f$ is: $$x z_y + y z_x - z = 0$$ **3.** - Given PDE: $$(y + z) p + (z + x) q = x + y$$ - Here $p = z_x$, $q = z_y$. - Use method of characteristics: $$\frac{dx}{y + z} = \frac{dy}{z + x} = \frac{dz}{x + y}$$ - Solve characteristic system: From $$\frac{dx}{y + z} = \frac{dy}{z + x}$$, cross multiply: $$(y + z) dy = (z + x) dx$$ - Rearranged: $$(y + z) dy - (z + x) dx = 0$$ - Similarly, from $$\frac{dy}{z + x} = \frac{dz}{x + y}$$, $$(z + x) dz = (x + y) dy$$ - Rearranged: $$(z + x) dz - (x + y) dy = 0$$ - These two differential forms can be integrated to find two independent integrals, say $F_1$ and $F_2$. - The general solution is an arbitrary function of these integrals: $$F(F_1, F_2) = 0$$ **Summary:** - Problem 1(a) PDE: $$z - x z_x - y z_y = (z_x)^2 (z_y)^2$$ - Problem 1(b) PDE: $$2 z = x z_x + y z_y$$ - Problem 2(a) PDE: $$y z_x - x z_y = 0$$ - Problem 2(b) PDE: $$x z_y + y z_x - z = 0$$ - Problem 3 general solution via characteristics as above.