Potential Function Exact Differential Equation Potential Example Contour of the

Potential function exact differential equation

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Potential Function Exact Differential Equation Potential Example Contour of the Potential Function Potential x y -5 -4 -3 -2 -1 0 1 2 3 4 5 -5 -4 -3 -2 -1 0 1 2 3 4 5 10 20 30 40 50 60 70 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Exact and Bernoulli Differentia — (13/26)
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Introduction Exact Differential Equations Bernoulli’s Differential Equation Gravity Potential Function Exact Differential Equation Exact Differential Equation Theorem Let the functions M , N , M y , and N x (subscripts denote partial derivatives) be continuous in a rectangular region R : α < x < β, γ < y < δ . Then the DE M ( x, y ) + N ( x, y ) y 0 = 0 is an exact differential equation in R if and only if M y ( x, y ) = N x ( x, y ) at each point in R . Furthermore, there exists a potential function φ ( x, y ) solving this differential equation with φ x ( x, y ) = M ( x, y ) φ y ( x, y ) = N ( x, y ) . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Exact and Bernoulli Differentia — (14/26)
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Introduction Exact Differential Equations Bernoulli’s Differential Equation Gravity Potential Function Exact Differential Equation Example Consider the differential equation 2 t cos( y ) + 2 + (2 y - t 2 sin( y )) y 0 = 0 Since ∂M ( t, y ) ∂y = - 2 t sin( y ) = ∂N ( t, y ) ∂t , this DE is exact Integrating Z (2 t cos( y ) + 2) dt = t 2 cos( y ) + 2 t + h ( y ) and Z (2 y - t 2 sin( y )) dy = y 2 + t 2 cos( y ) + k ( t ) It follows that the potential function is φ ( t, y ) = y 2 + 2 t + t 2 cos( y ) = C Joseph M. Mahaffy, h [email protected] i Lecture Notes – Exact and Bernoulli Differentia — (15/26)
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Introduction Exact Differential Equations Bernoulli’s Differential Equation Logistic Growth Equation Alternate Solution Bernoulli’s Equation Logistic Growth Equation 1 Logistic Growth Equation is one of the most important population models dP dt = rP 1 - P M , P (0) = P 0 This a 1 st order nonlinear differential equation It is separable, so can be written: Z dP P ( P M - 1 ) = - Z rdt = - rt + C Left integral requires partial fractions composition 1 P ( P M - 1 ) = A P + B ( P M - 1 ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Exact and Bernoulli Differentia — (16/26)
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Introduction Exact Differential Equations Bernoulli’s Differential Equation Logistic Growth Equation Alternate Solution Bernoulli’s Equation Logistic Growth Equation 2 Fundamental Theorem of Algebra gives A = - 1 and B = 1 /M , so integrals become Z (1 /M ) ( P M - 1 ) dP - Z dP P = - rt + C With a substitution, we have ln P ( t ) M - 1 - ln( P ( t )) = ln P ( t ) - M MP ( t ) = - rt + C Exponentiating (with K = e C ) P ( t ) - M MP ( t ) = Ke - rt or P ( t ) = M 1 - KMe - rt Joseph M. Mahaffy, h [email protected] i Lecture Notes – Exact and Bernoulli Differentia — (17/26)
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Introduction Exact Differential Equations Bernoulli’s Differential Equation Logistic Growth Equation Alternate Solution Bernoulli’s Equation Logistic Growth Equation 3 Logistic Growth Equation with initial condition is dP dt = rP 1 - P M , P (0) = P 0 With the initial condition and some algebra, the solution is P ( t ) = MP 0 P 0 + ( M - P 0 ) e - rt This solution took lots of work!
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