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# clase3handouts - Most differential equations have an innite...

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Most differential equations have an infinite number of solutions. In general, all solutions of a first-order DE form a family of solutions expressed with a single parameter C . Such a family is called the general solution . A member of the family that results from a specific value of C is called a particular solution. M.I. Bueno Differential Equations and Linear Algebra

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Example Find the general solution of dy dt = 3 y . Notice that y 0 is a solution. If y = 0 then dy y = 3 dt , dy y = 3 dt , log | y | = 3 t + c , | y | = e 3 t + c , | y | = e c e 3 t , | y | = De 3 t , D > 0 y = ± De 3 t . y = Ce 3 t , C R . M.I. Bueno Differential Equations and Linear Algebra
Example Find two particular solutions of dy dt = 3 y . Set of solutions: y = Ce 3 t , C R . For C = 0, we get the particular solution y 0. For C = 1, we get the particular solution y = e 3 t . M.I. Bueno Differential Equations and Linear Algebra

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Newton’s cooling law: Find the general solution of dT dt = k ( A - T ) , k > 0 , A - T < 0 . dT A - T = kdt , dT A - T = kdt , We know that f ( y ) f ( y ) dy = log | f ( y ) | + C . - - dT A - T = kdt , - log | A - T | = kt + C , - log ( T - A ) = kt + C , log ( T - A ) = - kt - C , T - A = e - kt - C , T = A + e - C e - kt , T = A + De - kt , D > 0 . M.I. Bueno Differential Equations and Linear Algebra
Initial-value problems We look for a solution to a differential equation that has a specified y -value y 0 at a given time t 0 . That specified point is called initial value . The combination of a first-order DE and an initial condition is called Initial-value problem . dy dt = f ( t , y ) , y ( t 0 ) = y 0 . While a DE generally has a family of solutions, an IVP usually has only one. M.I. Bueno Differential Equations and Linear Algebra

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Exercise Verify that y = e - t / 2 - e - 3 t satisfies the following initial-value problem: y + 3 y = e - t , y ( 0 ) = - 1 / 2 . Recall that d ( e u ) dt = du dt e u . y = - e - t / 2 - ( - 3 ) e - 3 t = - e - t / 2 + 3 e - 3 t . y + 3 y = ( - e - t / 2 + 3 e - 3 t ) + ( 3 e - t / 2 - 3 e - 3 t ) . = ( - 1 / 2 + 3 / 2 ) e - t = e - t . M.I. Bueno Differential Equations and Linear Algebra
Exercise Consider the second-order linear differential equation y - y - 2 y = 0. Verify that for any constants A and B , y = Ae 2 t + Be - t is a solution, taking into account that p = e 2 t and q = e - t are both solutions of the differential equation. y = Ap + Bq . y = Ap + Bq , y = Ap + Bq . y - y - 2 y = ( Ap + Bq ) - ( Ap + bq ) - 2 ( Ap + Bq ) , = A ( p - p - 2 p ) + B ( q - q - 2 q ) = A · 0 + B · 0 = 0 . M.I. Bueno Differential Equations and Linear Algebra

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Qualitative Analysis Studies properties of solutions directly from the differential equation itself. Direction fields is the most basic and useful tool of qualitative analysis for first-order DE.
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clase3handouts - Most differential equations have an innite...

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