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clase3handouts - Most differential equations have an...

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Unformatted text preview: 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 Example Find the general solution of dy dt = 3 y . Notice that y 0 is a solution. If y 6 = 0 then dy y = 3 dt , Z dy y = Z 3 dt , log | y | = 3 t + c , | y | = e 3 t + c , | y | = e c e 3 t , | y | = De 3 t , D > 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 Newtons cooling law: Find the general solution of dT dt = k ( A- T ) , k > , A- T < . dT A- T = kdt , Z dT A- T = Z kdt , We know that R f ( y ) f ( y ) dy = log | f ( y ) | + C .- Z- dT A- T = Z 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 > . 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 at a given time t . 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 ) = y . While a DE generally has a family of solutions, an IVP usually has only one. M.I. Bueno Differential Equations and Linear Algebra Exercise Verify that y = e- t / 2- e- 3 t satisfies the following initial-value problem: y + 3 y = e- t , y ( ) =- 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 00- 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 ....
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This note was uploaded on 03/14/2010 for the course MATH 3C taught by Professor Jacobs during the Fall '08 term at UCSB.

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clase3handouts - Most differential equations have an...

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