If dy-dx - e - 2y and y - 0 when x - 5- then value of x for y - 3 is

The correct option is D e^6 + 9/2 dy/dx=e^ 2y ∫ e^2ydy=∫ dx e^2y/2=x+c 2/2=5+c c= 9/2 x=e^6+9/2 ...

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Integrating Factor

If dy/dx = ...

Question

If $\frac{d y}{d x} = e^{- 2 y}$ and $y = 0$ when $x = 5$ , then value of x for $y = 3$ is

e^{5}

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e^{6} + 1

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\frac{e^{6} + 9}{2}

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l o g_{e} 6

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Solution

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Similar questions

\frac{d y}{d x} = e^{- 2 y}

y = 0

x = 5

x

y = 3

\frac{d y}{d x} + \frac{3}{{cos}^{2} x} y = \frac{1}{{cos}^{2} x}, x \in (\frac{- π}{3}, \frac{π}{3}),

y (\frac{π}{4}) = \frac{4}{3}

y (- \frac{π}{4})

x^{3} d y + x y d x = x^{2} d y + 2 y d x; y (2) = e

x > 1

y (4)

\frac{d y}{d x} = \frac{1 - cos x}{1 + cos x}

\frac{d y}{d x} = \sqrt{4 - y^{2}} (- 2 < y < 2)

\frac{d y}{d x} = 1 (y \neq 1)

{sec}^{2} x tan y d x + {sec}^{2} y tan x d y = 0

(e^{x} + e^{- x}) d y - (e^{x} - e^{- x}) d x = 0

\frac{d y}{d x} = (1 + x^{2}) (1 + y^{2})

y log y d x - x d y = 0

x^{4} \frac{d y}{d x} = - y^{5}

\frac{d y}{d x} = {sin}^{- 1} x

e^{x} tan y d x + (1 - e^{x}) {sec}^{2} y d y = 0

\frac{d y}{d x} = e^{- 2 y}, x = 5 \Rightarrow y = 0,

y = 3, f i n d t h e v a l u e o f 2 x

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