Mahaffy h jmahaffysdsuedu i Lecture Notes Introduction to Differential Equa

# Mahaffy h jmahaffysdsuedu i lecture notes

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Joseph M. Mahaffy, h [email protected] i Lecture Notes – Introduction to Differential Equa — (36/47) Subscribe to view the full document.

The Class — Overview The Class... Introduction Applications of Differential Equations Checking Solutions and IVP Evaporation Example Nonautonomous Example Introduction to Maple Evaporation Example 2 Solution: Show V ( t ) = (2 - 0 . 01 t ) 3 satisfies dV dt = - 0 . 03 V 2 / 3 , V (0) = 8 cm 3 V (0) = (2 - 0 . 01(0)) 3 = 8, so satisfies the initial condition Differentiate V ( t ), dV dt = 3(2 - 0 . 01 t ) 2 ( - 0 . 01) = - 0 . 03(2 - 0 . 01 t ) 2 But V 2 / 3 ( t ) = (2 - 0 . 01 t ) 2 , so dV dt = - 0 . 03 V 2 / 3 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Introduction to Differential Equations — (37/47) The Class — Overview The Class... Introduction Applications of Differential Equations Checking Solutions and IVP Evaporation Example Nonautonomous Example Introduction to Maple Evaporation Example 3 Solution (cont): Find the time of total desiccation Must solve V ( t ) = (2 - 0 . 01 t ) 3 = 0 Thus, 2 - 0 . 01 t = 0 or t = 200 It takes 200 days for complete desiccation Joseph M. Mahaffy, h [email protected] i Lecture Notes – Introduction to Differential Equa — (38/47) The Class — Overview The Class... Introduction Applications of Differential Equations Checking Solutions and IVP Evaporation Example Nonautonomous Example Introduction to Maple Evaporation Example 4 Graph of Desiccation 0 50 100 150 200 0 2 4 6 8 t (days) V(t) Volume of Water - V(t) = (2 - 0.01t) 3 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Introduction to Differential Equations — (39/47) The Class — Overview The Class... Introduction Applications of Differential Equations Checking Solutions and IVP Evaporation Example Nonautonomous Example Introduction to Maple Nonautonomous Example 1 Nonautonomous Example: Consider the nonautonomous differential equation with initial condition ( Initial Value Problem ): dy dt = - ty 2 , y (0) = 2 Show that the solution to this differential equation, including the initial condition, is y ( t ) = 2 t 2 + 1 Graph of the solution Joseph M. Mahaffy, h [email protected] i Lecture Notes – Introduction to Differential Equa — (40/47) The Class — Overview The Class... Introduction Applications of Differential Equations Checking Solutions and IVP Evaporation Example Nonautonomous Example Introduction to Maple Nonautonomous Example 2 Solution: Consider the solution y ( t ) = 2 t 2 + 1 = 2( t 2 + 1) - 1 The initial condition is y (0) = 2 0 2 + 1 = 2 Differentiate y ( t ), dy dt = - 2( t 2 + 1) - 2 (2 t ) = - 4 t ( t 2 + 1) - 2 However, - ty 2 = - t (2( t 2 + 1) - 1 ) 2 = - 4 t ( t 2 + 1) - 2 Thus, the differential equation is satisfied Joseph M. Mahaffy, h [email protected] i Lecture Notes – Introduction to Differential Equations — (41/47) The Class — Overview The Class... Introduction Applications of Differential Equations Checking Solutions and IVP Evaporation Example Nonautonomous Example Introduction to Maple Nonautonomous Example 3 Solution of Nonautonomous Differentiation Equation 0 1 2 3 4 5 0.5 1 1.5 2 t y(t) y = 2(t 2 + 1) -1 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Introduction to Differential Equa — (42/47) The Class — Overview The Class... Subscribe to view the full document. • Fall '08
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