If $y (x)$ is solution of differential equation satisfying \displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}+{\left(\frac{{{2}{x}+{1}}}{{x}}\right)}{y}={e}^{{-{2}{x}}},{y}

y (x)

is solution of differential equation satisfying

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

then (A)

y (lo g_{e} 2) = lo g_{e} 2

(B)

y (lo g_{e} 2) = \frac{lo g _{e} 2}{4}

(C)

y (x)

is decreasing is

(0, 1)

(D)

y (x)

is decreasing is

(\frac{1}{2}, 1)

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y (x)

is solution of differential equation satisfying \displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}+{\left(\frac{{{2}{x}+{1}}}{{x}}\right)}{y}={e}^{{-{2}{x}}},{y}

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Question Text

y (x)

is solution of differential equation satisfying \displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}+{\left(\frac{{{2}{x}+{1}}}{{x}}\right)}{y}={e}^{{-{2}{x}}},{y}

If $y (x)$ is solution of differential equation satisfying \displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}+{\left(\frac{{{2}{x}+{1}}}{{x}}\right)}{y}={e}^{{-{2}{x}}},{y}

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If y(x) is solution of differential equation satisfying \displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}+{\left(\frac{{{2}{x}+{1}}}{{x}}\right)}{y}={e}^{{-{2}{x}}},{y}

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Practice questions from similar books

If $y (x)$ is solution of differential equation satisfying \displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}+{\left(\frac{{{2}{x}+{1}}}{{x}}\right)}{y}={e}^{{-{2}{x}}},{y}