Let y = y(x) be the solution of the differential equation , x > 0 and y(e

Engineering

Mathematics

Introduction to Determinants

Question

Let y = y(x) be the solution of the differential equation

$(2 {xlog}_{e} x) \frac{dy}{dx} + 2 y = \frac{3}{x} \log_{e} x$ , x > 0 and y(e^–1) = 0. Then, y(e) is equal to

$- \frac{3}{e}$

$- \frac{3}{2 e}$

$- \frac{2}{e}$

$- \frac{2}{3 e}$

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$\frac{dy}{dx} + \frac{y}{xℓnx} = \frac{3}{2 x^{2}}$

$I.F. = e^{\int \frac{1}{xlnx} dx}$

= e^ℓn(ℓnx) = ℓnx

Solution is y. $lnx = \int (lnx) \frac{3}{2 x^{2}} dx + C$

Put ℓnx = t

$\frac{dx}{x} = dt$

$y \cdot lnx = \frac{3}{2} \int t . e^{- t} + c$

$= \frac{3}{2} [- {te}^{- t} - \int 1 \cdot (- e^{- t})) dt] + C = - \frac{3}{2} {te}^{- t} + \frac{3}{2} (- e^{- t}) + c$

$yℓnx = - \frac{3}{2} \frac{ℓnx}{x} - \frac{3}{2 x} + c$

Now $(\frac{1}{e}, 0) \Rightarrow c = 0$

when x = e

$y = - \frac{3}{e}$

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