Separation of Variables Modified Malthusian Growth Model Examples General

Separation of variables modified malthusian growth

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Separation of Variables Modified Malthusian Growth Model Examples General Separable Differential Equation General Separable Differential Equation 1 General Separable Differential Equation: An equivalent form is M ( t ) + N ( y ) dy dt = 0 (1) Theorem Let H 1 ( t ) and H 2 ( y ) be any antiderivatives of M and N with H 0 1 ( t ) = M ( t ) and H 0 2 ( y ) = N ( y ) . Then (1) can be written as a total derivative d dt [ H 1 ( t ) + H 2 ( y )] = 0 , so the general solution of (1) is H 1 ( t ) + H 2 ( y ) = C. Joseph M. Mahaffy, h [email protected] i Lecture Notes – Separable Differential Equations — (13/41) Introduction Separation of Variables Modified Malthusian Growth Model Examples General Separable Differential Equation General Separable Differential Equation 2 General Separable Differential Equation: M ( t ) + N ( y ) dy dt = 0 with H 0 1 ( t ) = M ( t ) and H 0 2 ( y ) = N ( y ) has an implicit representation of the solution H 1 ( t ) + H 2 ( y ) = C. Often this general form does not allow an explicit solution This form of solution should remind you of the total differential from Calculus M ( t ) dt + N ( y ) dy = 0 These result integral curves or level curves for each C This form will prove more useful for Exact DEs , which arise in potential problems from Physics Joseph M. Mahaffy, h [email protected] i Lecture Notes – Separable Differential Equations — (14/41) Introduction Separation of Variables Modified Malthusian Growth Model Examples General Separable Differential Equation Example 2 - Separable Differential Equation 1 Example 2: Consider the initial value problem dy dt = 4 sin(2 t ) y with y (0) = 1 Skip Example Solution: Begin by separating the variables, so Z y dy = 4 Z sin(2 t ) dt Solving the integrals gives y 2 2 = - 2 cos(2 t ) + C Joseph M. Mahaffy, h [email protected] i Lecture Notes – Separable Differential Equations — (15/41) Introduction Separation of Variables Modified Malthusian Growth Model Examples General Separable Differential Equation Example 2 - Separable Differential Equation 2 Solution (cont) Since y 2 2 = - 2 cos(2 t ) + C We write y 2 ( t ) = 2 C - 4 cos(2 t ) or y ( t ) = ± p 2 C - 4 cos(2 t ) From the initial condition y (0) = 1 = p 2 C - 4 cos(0) = 2 C - 4 Thus, 2 C = 5, and y ( t ) = p 5 - 4 cos(2 t ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Separable Differential Equations — (16/41)
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Introduction Separation of Variables Modified Malthusian Growth Model Examples General Separable Differential Equation Example 3 - Separable Differential Equation 1 Example 3: Consider the initial value problem dy dt = - y (1 + 2 t 2 ) t with y (1) = 2 Skip Example Solution: Begin by separating the variables, so Z dy y = - Z (1 + 2 t 2 ) t dt = - Z dt t - 2 Z t dt Solving the integrals gives ln( y ) = - ln( t ) - t 2 + C Joseph M. Mahaffy, h [email protected] i Lecture Notes – Separable Differential Equations — (17/41) Introduction Separation of Variables Modified Malthusian Growth Model Examples General Separable Differential Equation Example 3 - Separable Differential Equation 2 Solution (cont): Since ln( y ) = - ln( t ) - t 2 + C Exponentiate both sides to give y ( t ) = e - ln( t ) - t 2 + C = e - ln( t ) e - t 2 e C = A t e - t 2 where A = e C With the initial condition y (1) = 2 = A e - 1 or A = 2 e 1 The solution is y ( t ) = 2 t e 1 - t 2 Joseph M. Mahaffy,
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