Question

Form Partial differential eq. of u=ax+(1-a)y+b

Answer 1

Answer on Question #68366 – Math – Differential Equations

Question

Form Partial differential eq. of $u = ax + (1 - a)y + b$

Solution

We have relation

$$u = a x + (1 - a) y + b$$

Differentiate (1) first with respect to $x$ and then with respect to $y$

$$\frac{\partial u}{\partial x} = a \quad \text{or} \quad p = a$$ $$\frac{\partial u}{\partial y} = 1 - a \quad \text{or} \quad q = 1 - a$$

Substituting $a = p$ into $q = 1 - a$ we get the required partial equation

$$q = 1 - p,$$

hence

$$p + q = 1$$

or

$$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 1$$

**Answer**: The required partial equation is

$$p + q = 1$$

or

$$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 1$$

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