Let y = y(x) be the solution of the differential equation, , (x > 1).If 2y(2) = loge 4

Engineering

Mathematics

Linear Differential Equation

Question

Let y = y(x) be the solution of the differential equation, $x \frac{dy}{dx} + y = x \log_{e} x$ , (x > 1).

If 2y(2) = log_e 4 – 1, then y(e) is equal to

$\frac{e^{2}}{4}$

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

$\frac{e}{4}$

$\frac{- e}{2}$

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$\frac{dy}{dx} + \frac{1}{x} \cdot y = lnx$

$∴ I / F = e^{\int \frac{1}{x} dx} = e^{lnx} = x$

$∴ yx = \int x \cdot lnx dx = \frac{x^{2}}{2} lnx - \int \frac{1}{x} \cdot \frac{x^{2}}{2} dx$

$∴ xy = \frac{x^{2}}{2} lnx - \frac{x^{2}}{4} + c$

$Put x = 2 & y = \frac{1}{2} (\ln 4 - 1)$

⇒ ln 4 – 1 = 2 ln 2 – 1 + c $∴$ c = o

$∴ y = \frac{x}{2} lnx - \frac{x}{4}$

$\Rightarrow y = \frac{e}{2} - \frac{e}{4} = \frac{e}{4}$