# hw1f - (18) ( t ) y + 2 t y = ( t ) sin t t (19) d dt ( ( t...

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Math 2930 8/27/10 hw solutions problem to be graded - 18 2.1 16 y 0 + 2 t y = cos t t 2 , y ( π ) = 0 , t > 0 (1) Since it is a ﬁrst order linear equation use an integrating factor, μ ( t ) y 0 + 2 t μ ( t ) y = μ ( t ) cos t t 2 (2) d dt ( μ ( t ) y ) = μ ( t ) cos t t 2 (3) For (2) and (3) to hold, ( t ) dt = 2 μ ( t ) t μ ( t ) = C t 2 (4) Taking C = 1, (3) becomes, d dt ( t 2 y ) = cos t (5) Integrating both sides, Z ± d dt ( t 2 y ) ² dt = Z cos t dt (6) t 2 y = sin t + c (7) y = sin t + c t 2 (8) Applying the I.C. y ( π ) = 0 we ﬁnd c = 0, y = sin t t 2 (9) 17 y 0 - 2 y = e 2 t , y (0) = 2 (10) Following the solution to 16, 1

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μ ( t ) y 0 - 2 μ ( t ) y = μ ( t ) e 2 t (11) d dt ( μ ( t ) y ) = μ ( t ) e 2 t μ ( t ) = e - 2 t (12) d dt ( e - 2 t y ) = 1 (13) Z ± d dt ( e - 2 t y ) ² dt = Z dt (14) e - 2 t y = t + c (15) y = e 2 t ( t + c ) , y (0) = 2 c = 2 (16) y = e 2 t ( t + 2) (17) 18 t y 0 + 2 y = sin t, y ( π/ 2) = 1 , t >
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Unformatted text preview: (18) ( t ) y + 2 t y = ( t ) sin t t (19) d dt ( ( t ) y ) = ( t ) sin t t ( t ) = t 2 (20) d dt ( t 2 y ) = t sin t (21) Z d dt ( t 2 y ) dt = Z t sin t dt (22) t 2 y =-t cos t + Z cos t dt, ( integration by parts ) (23) =-t cos t + sin t + c (24) y = t-2 (-t cos t + sin t + c ) , y ( / 2) = 1 c = 2 4-1 (25) y = t-2 -t cos t + sin t + 2 4-1 (26) 2...
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## This note was uploaded on 01/21/2011 for the course MATH 2930 taught by Professor Terrell,r during the Spring '07 term at Cornell University (Engineering School).

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hw1f - (18) ( t ) y + 2 t y = ( t ) sin t t (19) d dt ( ( t...

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