Question

2y5-7y4 +12 y"' + 8 y" = 0

Answer 1

2y^3 -7y^4 + 12 y tprime + 8 y Prime = 0 Let y = e^mx be a solution of the above equation The auxiliary equation is, 2m^5 -7 m^4 + 12 m^3 + 8m^2 = 0 or, m^2 (2m^3 -7m^2 + 12m + 8) = 0 or, m^2 {2m^3 + m^2 -8m^2 -4m + 16 m + 8} = 0 or, m^2 {2m^2 (m + 1/2) -8m (m + 1/2) + 16 (m + 1/2) = 0 or, m^2 (m + 1/2) (2m^2 -8m + 16) = 0 or, m^2 (m + 1/2) (m^2 -4m + 8) = 0 Now, m^2 -4m + 8 = 0 = > m = 4 plusminus squareroot 16 -32/2 = 2 plusminus 2i So roots of the auxiliary equations are, 0, 0, 1/2, 2 + 2i, 2 -2i Hence the required solution is, Delta (x) = (a_1 + a_2 x) e^0.x + a_3 e^1/2 x + e^2x (a_4 cos 2x + a_5 sin 2x)

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