08Ma2aPrMidtermsol

08Ma2aPrMidtermsol - Math 2a Practical Fall 2008 Solutions...

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Unformatted text preview: Math 2a Practical, Fall 2008: Solutions to Midterm 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. (20pts) Solve the following differential equations. If you cannot find an antiderivative, leave the solution as an integral, if necessary. (a) y t + e t = y ( t 2 + 3) Solution. This is a first order linear ODE. We multiply both sides by t and rearrange to get y + te t = y ( t 3 + 3 t ) y + (- t 3- 3 t ) y =- te t The integrating factor is e R (- t 3- 3 t ) dt = e- 1 4 t 4- 3 2 t 2 . Hence we have [ e- 1 4 t 4- 3 2 t 2 y ] =- e- 1 4 t 4- 3 2 t 2 te t e- 1 4 t 4- 3 2 t 2 y = Z- te- 1 4 t 4- 3 2 t 2 + t dt + C y = e 1 4 t 4 + 3 2 t 2 ( Z- te- 1 4 t 4- 3 2 t 2 + t dt + C ) (b) y 00 + y = t 2 cos( t ) Solution. This is a second order linear ODE. First we solve the corresponding homogeneous equation by letting y = e rt y 00 + y = 0 ( r 2 + 1) e rt = 0 r 2 + 1 = 0 ( r- i )( r + i ) = 0 The roots to the characteristic equation are r = ± i , and so a fundamental set of solutions to the homogeneous equation is cos( t ), sin( t ), and the general solution to the homogeneous equation is y = C 1 cos( t ) + C 2 sin( t ) We find a particular solution using the method of undetermined coefficients. Since cos( t ) and sin( t ) are solutions to the homogeneous equation, as suggested by remark 4 on page 181, let Y ( t ) = ( At 3 + Bt 2 + Ct )cos( t ) + ( Dt 3 + Et 2 + Ft )sin( t ) So Y ( t ) = (3 At 2 +2 Bt + C )cos( t )- ( At 3 + Bt 2 + Ct )sin( t )+(3 Dt 2 +2 Et + F )sin( t )+( Dt 3 + Et 2 + Ft )cos( t ) = ( Dt 3 + (3 A + E ) t 2 + (2 B + F ) t + C )cos( t ) + (- At 3 + (- B + 3 D ) t 2 + (- C + 2 E ) t + F )sin( t ) and Y 00 ( t ) = (3 Dt 2 + (6 A + 2 E ) t + 2 B + F )cos( t )- ( Dt 3 + (3 A + E ) t 2 + (2 B + F ) t + C )sin( t ) +(- 3 At 2 + (- 2 B + 6 D ) t- C + 2 E )sin( t ) + (- At 3...
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08Ma2aPrMidtermsol - Math 2a Practical Fall 2008 Solutions...

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