# sol3_001 - Sample Problem Solutions for Exam Three 1 Solve...

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Unformatted text preview: Sample Problem Solutions for Exam Three 1. Solve the system x = x- 4 y ; y = x + y By the second equation, x = y- y , so x = y 00- y . Putting x = y- y into equation one, we get x = y- y- 4 y = y- 5 y . Then y 00- y = y- 5 y , so y 00- 2 y + 5 y = 0. This means y = e t ( c 1 cos 2 t + c 2 sin 2 t . Then y = e t ( c 1 cos 2 t + c 2 sin 2 t- 2 c 1 sin 2 t + 2 c 2 cos 2 t ), so x = y- y = e t (- 2 c 1 sin 2 t + 2 c 2 cos 2 t ). 2. Factor the differential equation y 00- 3 y +2 y = x into two first order equations and solve. ( D- 1)( D- 2) y = ( D- 1) z = x , where z = ( D- 2) y . Then z = e x R xe- x = e x (- x- 1) e- x =- x- 1, so y = e 2 x R (- x- 1) e- 2 x = e 2 x ( 1 2 x + 3 4 ) e- 2 x = 1 2 x + 3 4 x . 3. Find the general solution of y 000- 3 y + 2 y = 0. r 3- 3 r + 2 = ( r- 1)( r- 1)( r + 2), so y = c 1 e x + c 2 xe x + c 3 e- 2 x . 4. What does it mean to say that y 1 , y 2 and y 3 are independent? Show that x , x 2- 1 and x 2- 4 are independent using the definition. 5. Find the general solution of y ( viii )- y ( vii )- y ( iv ) + y ( iii ) = 0, given that r 8- r 7- r 4 + r 3 = r 3 ( r- 1) 2 ( r + 1)( r 2 + 1). This means that there is no non-trivial linear combination c 1 y 1 + c 2 y 2 + c 3 y 3 which is identically 0. Suppose c 1 x + c 2 ( x 2- 1) + c 3 ( x 2- 4) = 0. For x = 0, this gives- c 2- 4 c 3 = 0. For x = 1, this gives c 1- 3 c 3 = 0 For x =- 1, this gives- c 1- 3 c 3 = 0. Adding the last two equations, we see that c 3 = 0 and then that c 1 = 0 and by the first equation c 2 = 0 also. Thus x , x 2- 1 and x 2- 4 are linearly independent. 6. Use Undetermined Coefficients to solve y 000- y 00 + 4 y- 4 y = cosx . The auxiliary polynomial is r 3- r 2 + 4 r- 4 r = ( r- 1)( r 2 + 4), so the homogeneous solution is y h = c 1 e x + c 2 cos 2 x + c 3 sin 2 x . The particular solution is y p = Acos x + Bsin x . Then y p = Bcos x- Asin x , y 00 p =- Acos x- Bsin x and y 000 p =- Bcos x + Asin x . Thus L [ y p ] = (- 3 A + 3 B ) cos x + (- 3 A- 3 B ) sin x = cos x . So- 3 A + 3 B = 1 and- 3 A- 3 B = 0, which means A =- 1 6 and B = 1 6 ....
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sol3_001 - Sample Problem Solutions for Exam Three 1 Solve...

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