HW2solutions

# HW2solutions - Differential Equations Assignment...

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Differential Equations Assignment 2: (Solutions) Quantitative Methods for First Order Equations and Modeling: 1. Use the method of integrating factors to find the general solution to the following first order differential equations: (a) y + 3 y = 0 The integrating factor is given by μ = e 3 t . Multiplying by this factor we find: e 3 t y + 3 e 3 t y = 0 . Applying the product rule in reverse we know: d dt e 3 t y = 0 . Integrating with respect to t and solving for y we obtain the general solution. d dt e 3 t y dt = 0 dt, e 3 t y = C, y = Ce - 3 t . (b) ty + 2 y = 1 t First we write the equation in the proper form for applying the integrating factor formula y + 2 t y = 1 t 2 Now we find the integrating factor. μ = e R 2 t dt = e 2 ln( t ) = e ln( t 2 ) = t 2 . Multiplying by this integrating factor: t 2 y + 2 ty = 1 . 1

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Applying the product rule in reverse we know: d dt t 2 y = 1 . Integrating with respect to t and solving for y we obtain the general solution. d dt t 2 y dt = 1 dt, t 2 y = t + C, y = 1 t + C t 2 . (c) y - y = e - t The integrating factor for this equation is μ = e - t . Multiplying by this factor we obtain: e - t y - e - t y = e - 2 t . Using the product rule in reverse we know: d dt e - t y = e - 2 t . Integrating with respect to t and solving for y we obtain the general solution. d dt e - t y dt = e - 2 t dt, e - t y = - 1 2 e - 2 t + C, y = - 1 2 e - t + Ce t . (d) y + cot( t ) y = 0 For this problem, we sidestep the calculation of the integrating factor and take advantage of the structure of the equation. Using the definition of cot( t ) we know: y + cos( t ) sin( t ) y = 0 2
Multiplying by sin( t ) , we find that we already have the correct form to apply the method of integrating factors: sin( t ) y + cos( t ) y = 0 d dt [sin( t ) y ] = 0 Integrating with respect to t and solving for y we obtain the general solution. d dt [sin( t ) y ] dt = 0 sin( t ) y = C y = C sin( t ) = C csc( t ) 2. Solve the following initial value problems using any appropriate technique: (a) y + 4 y = e - 2 t , y (0) = 1, This is a linear equation, so the method of integrating factors will work. The integrating factor is μ = e 4 t . Multiplying by this factor we get: e 4 t y + 4 e 4 t y = e 2 t Applying the product rule in reverse we find: d dt e 4 t y = e 2 t . Integrating and solving we get: d dt e 4 t y dt = e 2 t dt. e 4 t y = 1 2 e 2 t + C y = 1 2 e - 2 t + Ce - 4 t Substituting the initial condition we find: 1 = 1 2 + C C = 1 2 3

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Thus, the particular solution to this initial value problem becomes: y = 1 2 e - 2 t + 1 2 e - 4 t Since this solution contains only exponentials of the form e - at , each term will go to zero and y 0 and t → ∞ . (b) y = t 1+ y 2 y , y (1) = - 1 This equation is non-linear so we try to apply the method of separation. Sepa- rating all of the y terms we get: yy 1 + y 2 = t We now attempt to integrate the equation in this separated form: yy 1 + y 2 dt = tdt.
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