Answer on Question #68368 – Math – Differential Equations
Question
Form Partial differential eq. of 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 \begin{array}{l}
u = x y + y \sqrt {x ^ {2} - a ^ {2}} + b \\
p = \frac {\partial u}{\partial x} = y + \frac {x y}{\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 {x y}{q - x} \\
\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
A partial differential equation (PDE) is
p = y + x y q − x , p = y + \frac {x y}{q - x}, p = y + q − x x y ,
that is,
∂ u ∂ x = y + x y ∂ u ∂ y − x \frac {\partial u}{\partial x} = y + \frac {x y}{\frac {\partial u}{\partial y} - x} ∂ x ∂ u = y + ∂ y ∂ u − x x y
Answer: p = y + x y q − x p = y + \frac{xy}{q - x} p = y + q − x x y or ∂ 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 .
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