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08Ma2aPracSol1

# 08Ma2aPracSol1 - Math 2a Practical Fall 2008 Solutions to...

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Math 2a Practical, Fall 2008: Solutions to Homework 1 You are bound by the honor code to not look at this file if you did not take Math 2a in the Fall of 2008 1. (12 points) For each of the following differential equations, say what its order is, whether it is an ODE or a PDE, and whether it is linear or nonlinear. Solution. (a) t 2 d 2 y dt 2 + t dy dt + 2 y = sin( t ) Order 2 linear ODE (b) d 3 u dx 3 + d 2 u dx 2 + du dx 3 = u x Order 3 non-linear ODE (c) - ∂V ∂t = 2 V ∂x 2 + 2 V ∂y 2 Order 2 linear PDE (d) dy dt + sin( t + y ) = 0 Order 1 non-linear ODE 2. (15 points) Find the solution of the following equations or initial value problems as explicitly as you can: (a) y 0 - y = 2 te 2 t , y (0) = 1 Solution. This is a linear first order ODE. We find the integrating factor μ ( t ) = e R - 1 dt = e - t Multiplying both sides of the equation by the integrating factor e - t gives e - t y 0 - e - t y = 2 te t d dt ( e - t y ) = 2 te t Integrating both sides (you can integrate the right hand side using integration by parts) gives e - t y = Z 2 te t dt

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e - t y = 2 te t - 2 e t + C y = 2 te 2 t - 2 e 2 t + Ce t Plugging in the initial value y (0) = 1 gives 1 = 0 - 2 + C C = 3. So y = 2 te 2 t - 2 e 2 t + 3 e t (b) y 0 + 2 t y = cos( t ) t 2 , y ( π ) = 0 , t > 0 Solution. This is a linear first order ODE. We find the integrating factor μ ( t ) = e R 2 t dt = e 2 ln | t | = t 2 Multiplying both sides of the equation by the integrating factor t 2 gives t 2 y 0 + 2 ty = cos( t ) d dt ( t 2 y ) = cos( t ) Integrating both sides gives t 2 y = sin( t ) + C y = sin( t ) + C t 2 Plugging in the initial value gives C=0, so y = sin( t ) t 2 (c) t 3 y 0 + 4 t 2 y = e - t Solution.
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08Ma2aPracSol1 - Math 2a Practical Fall 2008 Solutions to...

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