Quiz5Bsolution

# Quiz5Bsolution - Find the solution of the given initial...

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MATH 415.01 QUIZ 5B Feb 10, 2010 SHOW ALL YOUR WORK! GOOD LUCK! NAME: 1. (6.5 pts.) Use the method of reduction of order to ﬁnd a second solution to the following diﬀerential equation. xy 00 - y 0 + 4 x 3 y = 0 , x > 0; y 1 ( x ) = sin x 2 Let y ( x ) = v ( x ) sin x 2 . Then, y 0 ( x ) = sin x 2 v 0 ( x )+2 x cos x 2 v ( x ) and y 00 ( x ) = sin x 2 v 00 ( x )+4 x cos x 2 v 0 ( x )+(2 cos x 2 - 4 x 2 sin x 2 ) v ( x ) Plug this into the equation above: x (sin x 2 v 00 ( x ) + 4 x cos x 2 v 0 ( x ) + (2 cos x 2 - 4 x 2 sin x 2 ) v ( x )) - (sin x 2 v 0 ( x ) + 2 x cos x 2 v ( x )) + 4 x 3 sin x 2 v = 0 i.e. x sin x 2 v 00 ( x ) + 4 x 2 cos x 2 v 0 ( x ) + (2 x cos x 2 - 4 x 3 sin x 2 ) v ( x ) - sin x 2 v 0 ( x ) - 2 x cos x 2 v ( x ) + 4 x 3 sin x 2 v = 0 after cancellation, x sin x 2 v 00 ( x ) + 4 x 2 cos x 2 v 0 ( x ) - sin x 2 v 0 ( x ) = 0 i.e. v 00 ( x ) + (4 x cos x 2 sin x 2 - 1 x ) v 0 ( x ) = 0 Let u ( x ) = v 0 ( x ) then rewrite the equation as u 0 ( x ) + (4 x cos x 2 sin x 2 - 1 x ) u ( x ) = 0 i.e. du dx + (4 x cos x 2 sin x 2 - 1 x ) u = 0 Use seperation of variables to solve this equation: du u = - (4 x cos x 2 sin x 2 - 1 x ) dx ln( u ) = Z du u = - 2 Z 2 x cos x 2 sin x 2 dx + Z 1 x dx = - 2 ln(sin x 2 )+ln( x ) = ln(sin x 2 ) - 2 +ln( x ) = ln( x (sin x 2 ) - 2 ) Thus, the solution is ln u = ln x (sin x 2 ) 2 i.e. u = x (sin x 2 ) 2 . Then substitute v 0 back in the equation and solve for v ( x ) v ( x ) = Z v 0 dx = Z x (sin x 2 ) 2 dx = - 1 2 cot x 2 Thus, y = v ( x ) y 1 ( x ) = - 1 2 cot x 2 sin x 2 = - 1 2 cos x 2 i.e. y 2 ( x ) = cos x 2

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2. (6 pts.)
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Unformatted text preview: Find the solution of the given initial value problem below. y 00 + y-2 y = 2 t, y (0) = 0 , y (0) = 1 Charactheristic equation: r 2 + r-2 = 0 i.e. roots are -2 and 1. Then, solution of the homogenous part is y = c 1 e-2 t + c 2 e t Method of undetermined coeﬃcients: Y = At + B , Y = A , Y 00 = 0. Plug these in the equation above, A-2( At + B ) = 2 t i.e. A-2 B-2 At = 2 t . So, A =-1 and B =-1 / 2. General solution becomes y = c 1 e-2 t + c 2 e t-t-1 / 2 . Now using the initial conditions, y (0) = c 1 + c 2-1 / 2 = 0 and y (0) =-2 c 1 + c 2-1 = 1 we get c 1 =-1 / 2 and c 2 = 1. y =-1 / 2 e-2 t + e t-t-1 / 2 ....
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## This note was uploaded on 07/26/2011 for the course MATH 415 taught by Professor Costin during the Spring '07 term at Ohio State.

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Quiz5Bsolution - Find the solution of the given initial...

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