# secnhom_a - EXAMPLES OF NON-HOMOGENEOUS SECOND ORDER LINEAR...

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Unformatted text preview: EXAMPLES OF NON-HOMOGENEOUS SECOND ORDER LINEAR EQUATIONS ANSWERS In each of the following, we will NOT solve the homogeneous equation. We solve the non-homogeneous problem, give the general solution, and then solve for initial conditions if required. 1 . d 2 y dx 2- dy dx- 2 y = x Trial Solution: y ( x ) = A x + B . y = A, y 00 = 0. Substituting, 0- A- 2 B- 2 A x = x so, equating coefficients of like terms,- 2 A = 1 ⇒ A =- 1 / 2, and A + 2 B = 0 ⇒ 2 B = 1 / 2 or B = 1 / 4. Hence: (a) Particular Solution: y ( x ) = (- 1 / 2) x + (1 / 4). (b) General Solution: y ( x ) = C 1 e- t + C 2 e 2 t- ( x/ 2) + 1 / 4 2 . d 2 y dx 2- dy dx- 2 y = 3 x 2 Trial Solution: y ( x ) = A x 2 + B x + C . y = 2 A x + B, y 00 = 2 A . Substituting,- 2 A x 2 + (- 2 B- 2 A ) x + (2 A- B- 2 C ) = 3 x 2 so, equating coefficients of like terms:- 2 A = 3 ,- 2 A- 2 B = 0, and 2 A- B- 2 C = 0 = ⇒ A = (- 3 / 2) , B = (3 / 2) and C = (- 9 / 4). Hence: (a) Particular Solution: y ( x ) = (- 3 / 2) x 2 + (3 / 2) x- (9 / 4). (b) General Solution: y ( x ) = C 1 e- x + C 2 e 2 x- (3 / 2) x 2 + (3 / 2) x- (9 / 4). 3 . y 00- 16 y = sin (2 x ) , with the initial conditions y (1) = 1 y (1) = 0 , given that the general solution of the homogeneous problem is y h ( x ) = c 1 e- 4 x + c 2 e 4 x . Trial Solution: y ( x ) = A cos (2 x ) + B sin (2 x ) , y =- 2 A sin (2 x ) + 2 B cos (2 x ) , y 00 =- 4 A cos (2 x )- 4 B sin (2 x ). Substituting,- 4 A cos (2 x )...
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secnhom_a - EXAMPLES OF NON-HOMOGENEOUS SECOND ORDER LINEAR...

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