Solving a Linear DE 2 With the integrating factor our example can be write e 2

Solving a linear de 2 with the integrating factor our

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Solving a Linear DE 2 With the integrating factor , our example can be write e - 2 t dy dt - 2 e - 2 t y = d dt e - 2 t y = (4 - t ) e - 2 t . The quantity d dt e - 2 t y is a total derivative, so we integrate both sides giving: e - 2 t y ( t ) = Z (4 - t ) e - 2 t dt + C = 1 4 (2 t - 7) e - 2 t + C, so y ( t ) = 1 4 (2 t - 7) + Ce 2 t . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Linear Differential Equations — (17/64)
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Introduction Falling Cat 1 st Order Linear DEs Examples DE with Only Time Varying Function Linear Differential Equation Integrating Factor General Solution 1 st Order Linear DE General Integrating Factor The differential equation for the integrating factor is ( t ) dt = p ( t ) μ ( t ) or 1 μ ( t ) ( t ) dt = p ( t ) . Note that d (ln( μ ( t ))) dt = 1 μ ( t ) ( t ) dt . It follows that ln( μ ( t )) = Z p ( t ) dt. The general integrating factor satisfies μ ( t ) = e R p ( t ) dt . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Linear Differential Equations — (18/64)
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Introduction Falling Cat 1 st Order Linear DEs Examples DE with Only Time Varying Function Linear Differential Equation Integrating Factor General Solution 1 st Order Linear DE 1 st Order Linear DE Solution Thus, the 1 st Order Linear DE Solution dy dt + p ( t ) y = g ( t ) with μ ( t ) = e R p ( t ) dt is integrated to produce μ ( t ) y ( t ) = Z μ ( t ) g ( t ) dt + C. Theorem (Solution of 1 st Order Linear DE) With the 1 st Order Linear DE given above and assuming integrability of p ( t ) and g ( t ) , then the solution is given by y ( t ) = e - R p ( t ) dt Z e R p ( t ) dt g ( t ) dt + C . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Linear Differential Equations — (19/64)
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Introduction Falling Cat 1 st Order Linear DEs Examples DE with Only Time Varying Function Linear Differential Equation Integrating Factor General Solution 1 st Order Linear DE Linear DE –Example 1 Consider the Linear DE Solution t dy dt - y = 3 t 2 sin( t ) . 1. Put this equation into standard form, so divide by t and obtain dy dt - 1 t y = 3 t sin( t ) . (3) 2. Observe p ( t ) = - 1 t , so find integrating factor μ ( t ) = e R ( - 1 /t ) dt = e - ln( t ) = 1 t . 3. Multiply (3) by μ ( t ) giving 1 t dy dt - 1 t 2 y = d dt y t = 3 sin( t ) . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Linear Differential Equations — (20/64)
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Introduction Falling Cat 1 st Order Linear DEs Examples DE with Only Time Varying Function Linear Differential Equation Integrating Factor General Solution 1 st Order Linear DE Linear DE –Example 1 The previous slide showed the transformation of t dy dt - y = 3 t 2 sin( t ) with the integrating factor μ ( t ) = 1 t to d dt y t = 3 sin( t ) . 4. Integrate this equation 1 t y ( t ) = 3 Z sin( t ) dt + C = - 3 cos( t ) + C, which gives the solution y ( t ) = - 3 t cos( t ) + Ct. Joseph M. Mahaffy, h [email protected] i Lecture Notes – Linear Differential Equations — (21/64)
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Introduction Falling Cat 1 st Order Linear DEs Examples Pollution in a Lake Example 2 Mercury in Fish Modeling Mercury in Fish Pollution in a Lake 1 Pollution in a Lake: Introduction An urgent problem in modern society is how to reduce pollution in our water sources These are complex issues, requiring a multidisciplinary approach, and are often politically intractable because of the key role that
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