Answer on Question #68401 – Math – Differential Equations
Question
Obtain the partial differential equation by eliminating the arbitrary constant from the relation
u = x y + y x 2 − a 2 + b u = xy + y\sqrt{x^2 - a^2} + b u = x y + y x 2 − a 2 + b
Solution
u = x y + y x 2 − a 2 + b p = ∂ u ∂ x = y + x y x 2 − a 2 q = ∂ u ∂ y = x + x 2 − a 2 x 2 − a 2 = q − x → p = y + x y q − x ∂ u ∂ x = y + x y ∂ u ∂ y − x is a partial differential equation. \begin{array}{l}
u = xy + y\sqrt{x^2 - a^2} + b \\
p = \frac{\partial u}{\partial x} = y + \frac{xy}{\sqrt{x^2 - a^2}} \\
q = \frac{\partial u}{\partial y} = x + \sqrt{x^2 - a^2} \\
\sqrt{x^2 - a^2} = q - x \rightarrow p = y + \frac{xy}{q - x} \\
\frac{\partial u}{\partial x} = y + \frac{xy}{\frac{\partial u}{\partial y} - x} \text{ is a partial differential equation.}
\end{array} u = x y + y x 2 − a 2 + b p = ∂ x ∂ u = y + x 2 − a 2 x y q = ∂ y ∂ u = x + x 2 − a 2 x 2 − a 2 = q − x → p = y + q − x x y ∂ x ∂ u = y + ∂ y ∂ u − x x y is a partial differential equation.
Answer: ∂ u ∂ x = y + x y ∂ u ∂ y − x \frac{\partial u}{\partial x} = y + \frac{xy}{\frac{\partial u}{\partial y} - x} ∂ x ∂ u = y + ∂ y ∂ u − x x y .
Answer provided by https://www.AssignmentExpert.com