The solution of the ordinary differential equation dydx-2y-0 for the boundary condition- y - 5 at x - 1 is

Given dydx+2y=0 and at x = 1, y = 5 Differential equation is, (D + 2)y = 0 ⇒ D = 2 Its solution is nbsp; y = C_1e^ 2x At x = 1, y = 5 ⇒ 5 =C_1e^ 2 ⇒ ...

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Engineering Mathematics

Homogeneous Linear Differential Equations (General Form of LDE)

The solution ...

Question

The solution of the ordinary differential equation $\frac{d y}{d x} + 2 y = 0$ for the boundary condition, y = 5 at x = 1 is

y = e^{- 2 x}

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y = 2 e^{- 2 x}

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y = 10.95 e^{- 2 x}

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y = 36.95

e^{- 2 x}

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Solution

The correct option is D y = 36.95 $e^{- 2 x}$
Given $\frac{d y}{d x} + 2 y = 0$

and at x = 1, y = 5

Differential equation is,

(D + 2)y = 0

$\Rightarrow D = - 2$

Its solution is

y = $C_{1} e^{- 2 x}$

At x = 1, y = 5

$\Rightarrow 5 = C_{1} e^{- 2}$

$\Rightarrow 5 e^{2} = C_{1}$

= 36.95

Hence required solution is

y = 36.95 $e^{- 2 x}$

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The solution of the ordinary differential equation dydx+2y=0 for the boundary condition, y = 5 at x = 1 is

The correct option is D y = 36.95e−2xGiven dydx+2y=0 and at x = 1, y = 5 Differential equation is, (D + 2)y = 0 ⇒D=−2 Its solution is y = C1e−2x At x = 1, y = 5 ⇒5=C1e−2 ⇒5e2=C1 = 36.95 Hence required solution is y = 36.95e−2x

The solution of the ordinary differential equation $\frac{d y}{d x} + 2 y = 0$ for the boundary condition, y = 5 at x = 1 is