The soluitons of the differential equations d2ydx2-2dydx-2y-0 are

Given (D^2+2D+2)y=0 m= 1± i So nbsp; nbsp;y=e^ x[C_1 cos x+C_2 sin x] or C_1e^( 1+i)x+C_2e^( 1 i)x=C_1e^ (1 i)x+C_2e^ (1+i)x Herer e^ (1 i)x and e^ (1+i)x ...

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Engineering Mathematics

Homogeneous Linear Differential Equations (General Form of LDE)

The soluitons...

Question

The soluitons of the differential equations $\frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} + 2 y = 0$ are

e^{- (1 + i) x}, e^{- (1 - i) x}

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e^{(1 + i) x}, e^{(1 - i) x}

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e^{- (1 + i) x}, e^{(1 - i) x}

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e^{(1 + i) x}, e^{- (1 - i) x}

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Solution

The correct option is A $e^{- (1 + i) x}, e^{- (1 - i) x}$
Given $(D^{2} + 2 D + 2) y = 0$
$m = - 1 \pm i$
So $y = e^{- x} [C_{1} c o s x + C_{2} s i n x]$ or $C_{1} e^{(- 1 + i)} x + C_{2} e^{(- 1 - i)} x = C_{1} e^{- (1 - i) x} + C_{2} e^{- (1 + i) x}$
Herer $e^{- (1 - i) x}$ and $e^{- (1 + i) x}$ are independent solutions.

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The soluitons of the differential equations d2ydx2+2dydx+2y=0 are

The correct option is A e−(1+i)x,e−(1−i)xGiven (D2+2D+2)y=0 m=−1±i So y=e−x[C1cosx+C2sinx] or C1e(−1+i)x+C2e(−1−i)x=C1e−(1−i)x+C2e−(1+i)x Herer e−(1−i)x and e−(1+i)x are independent solutions.

The soluitons of the differential equations $\frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} + 2 y = 0$ are