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Solve the $(D^4 + 4)y = 0$.

written 9.5 years ago by teamques10 ★ 70k

modified 3.9 years ago by phenjoisilab • 0

engineering mathematics

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written 9.5 years ago by teamques10 ★ 70k

Auxiliary equation is $D^4 + 4 = 0$

Adding and subtracting $4D^2$

$D^4 + 4D^2 + 4 – 4D^2 = 0$

$(D^2 + 2)^2 – (2D)^2 = 0$

$(D^2 + 2 + 2D) (D^2 + 2 - 2D) = 0$

$D^2 + 2D + 2 = 0 or D^2 – 2D + 2 = 0$

$\therefore D=-1\pm i$ or $D=1\pm i$

$\therefore C.f\space is\space y_c=e^{-x}(C_1\cos x + C_2\sin x)+e^x(C_3\cos x+C_4\sin x)$

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