The general solution of the differential equation d2ydx2-2dydx-5y-0 in terms of arbitrary constant K1 and K2 is

D^2ydx^2+2dydx 5y=0 ⇒ (D^2+2D 5)y=0 Auxiliary equation is m^2+2m 5=0 m= 1±√(6) So, in terms of arbitary constants k_1 and k_2 we get, y=k_1e^m_1x+k_2e^m_2x y=k_ ...

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Engineering Mathematics

Homogeneous Linear Differential Equations (General Form of LDE)

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Question

The general solution of the differential equation $\frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - 5 y = 0$ in terms of arbitrary constant $K_{1}$ and $K_{2}$ is

K_{1} e^{(- 1 + \sqrt{6})} x + K_{2} e^{(- 1 - \sqrt{6})} x

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K_{1} e^{(- 1 - \sqrt{8})} x + K_{2} e^{(- 1 - \sqrt{8})} x

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K_{1} e^{(- 2 - \sqrt{6})} x + K_{2} e^{(- 2 - \sqrt{6})} x

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K_{1} e^{(- 2 + \sqrt{8})} x + K_{2} e^{(- 2 - \sqrt{8})} x

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Solution

The correct option is A $K_{1} e^{(- 1 + \sqrt{6})} x + K_{2} e^{(- 1 - \sqrt{6})} x$
$\frac{d^{2} y}{d x^{2}} + \frac{2 d y}{d x} - 5 y = 0$ $\Rightarrow$ $(D^{2} + 2 D - 5) y = 0$
Auxiliary equation is $m^{2} + 2 m - 5 = 0$
$m = - 1 \pm \sqrt{6}$
So, in terms of arbitary constants $k_{1}$ and $k_{2}$ we get,
$y = k_{1} e^{m_{1} x} + k_{2} e^{m_{2} x}$
$y = k_{1} e^{(} - 1 + \sqrt{6} x) + k_{2} e^{(} - 1 - \sqrt{6} x)$

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The general solution of the differential equation d2ydx2+2dydx−5y=0 in terms of arbitrary constant K1 and K2 is

The correct option is A K1e(−1+√6)x+K2e(−1−√6)xd2ydx2+2dydx−5y=0 ⇒ (D2+2D−5)y=0 Auxiliary equation is m2+2m−5=0 m=−1±√6 So, in terms of arbitary constants k1 and k2 we get, y=k1em1x+k2em2x y=k1e(−1+√6x)+k2e(−1−√6x)

The general solution of the differential equation $\frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - 5 y = 0$ in terms of arbitrary constant $K_{1}$ and $K_{2}$ is