Gravity Potential Function Exact Differential Equation Gravity Potential

# Gravity potential function exact differential

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Gravity Potential Function Exact Differential Equation Gravity Potential Function Consider a potential function , φ ( x, y ) By implicit differentiation ( x, y ) dx = ∂φ ∂x + ∂φ ∂y dy dx If the potential function satisfies φ ( x, y ) = C ( level potential field ), then ( x, y ) dx = 0 This gives rise to an Exact differential equation Joseph M. Mahaffy, h [email protected] i Lecture Notes – Exact and Bernoulli Differentia — (7/26)

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Introduction Exact Differential Equations Bernoulli’s Differential Equation Gravity Potential Function Exact Differential Equation Gravity Definition Suppose there is a function φ ( x, y ) with ∂φ ∂x = M ( x, y ) and ∂φ ∂y = N ( x, y ) . The first-order differential equation given by M ( x, y ) + N ( x, y ) dy dx = 0 is an exact differential equation with the implicit solution satisfying: φ ( x, y ) = C. Joseph M. Mahaffy, h [email protected] i Lecture Notes – Exact and Bernoulli Differentia — (8/26)
Introduction Exact Differential Equations Bernoulli’s Differential Equation Gravity Potential Function Exact Differential Equation Example 1 Example: Consider the differential equation: (2 x + y cos( xy )) + (4 y + x cos( xy )) dy dx = 0 This equation is clearly nonlinear and not separable. We hope that it might be exact ! If it is exact , then there must be a potential function , φ ( x, y ) satisfying: ∂φ ∂x = 2 x + y cos( xy ) and ∂φ ∂y = 4 y + x cos( xy ) . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Exact and Bernoulli Differentia — (9/26)

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Introduction Exact Differential Equations Bernoulli’s Differential Equation Gravity Potential Function Exact Differential Equation Example 2 Example (cont): Begin with ∂φ ∂x = M ( x, y ) = 2 x + y cos( xy ) . Integrate this with respect to x , so φ ( x, y ) = Z (2 x + y cos( xy )) dx = x 2 + sin( xy ) + h ( y ) , where h ( y ) is some function depending only on y Similarly, we want ∂φ ∂y = N ( x, y ) = 4 y + x cos( xy ) . Integrate this with respect to y , so φ ( x, y ) = Z (4 y + x cos( xy )) dy = 2 y 2 + sin( xy ) + k ( x ) , where k ( x ) is some function depending only on x Joseph M. Mahaffy, h [email protected] i Lecture Notes – Exact and Bernoulli Differentia — (10/26)
Introduction Exact Differential Equations Bernoulli’s Differential Equation Gravity Potential Function Exact Differential Equation Example 3 Example (cont): The potential function , φ ( x, y ) satisfies φ ( x, y ) = x 2 +sin( xy )+ h ( y ) and φ ( x, y ) = 2 y 2 +sin( xy )+ k ( x ) for some h ( y ) and k ( x ) Combining these results yields the solution φ ( x, y ) = x 2 + 2 y 2 + sin( xy ) = C. Implicit differentiation yields: dx = (2 x + y cos( xy )) + (4 y + x cos( xy )) dy dx = 0 , the original differential equation. Joseph M. Mahaffy, h [email protected] i Lecture Notes – Exact and Bernoulli Differentia — (11/26)

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Introduction Exact Differential Equations Bernoulli’s Differential Equation Gravity Potential Function Exact Differential Equation Potential Example Graph of the Potential Function -4 -2 0 2 4 -4 -2 0 2 4 0 10 20 30 40 50 60 70 80 x Potential y Potential 0 10 20 30 40 50 60 70 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Exact and Bernoulli Differentia — (12/26)
Introduction Exact Differential Equations Bernoulli’s Differential Equation Gravity

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