6/6/2018

4y'' + 9y = 15 Let the solution be y = yh + yp where yh is the solution of homogeneous equation which is 4y''+9y = 0 yp is the particular solution of 4y''+9y = 15 Solving yh: 4y'' + 9y = 0 The characterstic equation is 4x^2 + 9 = 0 => x = +(3/2)i, -(3/2)i So, the solution is yh = c1cos((3/2)x) + c2sin((3/2)x) Solving yp: 4y'' + 9y = 15 So, yp'' + 9yp = 15 Let yp = Ax^2 + Bx + C yp' = 2Ax + B yp'' = 2A => 8A + Ax^2 + Bx + C = 15 => A = 0, B = 0 and C = 15 So, C = 15 So, yp = 15 So, the solution is y = yh + yp => y = c1cos((3/2)x) + c2sin((3/2)x) + 15 The solution for 4y'' + 9y = 15 is y = c1cos((3/2)x) + c2sin((3/2)x) + 15 The associated homogenous equation is 4y�� + 9y = 0 4y''+9y = 0 y1 = (3/2)i , y2 = -(3/2)i yh = c1cos(3x/2) + c2sin (3x/2) since g(x) is a constant, we assume yp = A yp� = 0, yp�� = 0 https://www.textsheet.com/retort

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6/6/2018

Substituting this in the differential equation gives: 0 + 9A = 15. This gives A = 15/9 = 5/3. yp = 5/3 The general solution of the differential equation is y = c1cos(3x/2) + c2sin(3x/2) + 5/3

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