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TS EAMCET 04 May 2019 Shift 2 Solved Paper

Section: Mathematics

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Question : 80 of 160

Marks: +1, -0

If x log x

\frac{dy}{dx} + y

\log x^2

and y (e) = 0 , then y

(e^2)

=

0
1
$\frac{1}{2}$
$\frac{3}{2}$

Solution:

Consider the differential equation.

x \log x \; \frac{dy}{dx} + y \;\; = \log x^2

\; \frac{dy}{dx} + \; \frac{y}{x \log x} \;\; = \; \frac{2}{x}

Calculate the integrating factor.

IF \;\; = e^{\int \; \frac{1}{x \log x} dx}

\;\; = e^{\log (\log x)}

\;\; = \log x

Therefore,

y \log x = \int \; \frac{2 \log x}{x} dx + c

y \log x = (\log x)^2 + x

\; \text{ At } \; x = e, y = 0

0 = 1 + c

c = -1

Therefore,

y \log x = (\log x)^2 - 1

At

x = e^2

y \log e^2 \;\; = (\log e^2)^2 - 1

2y \;\; = 4 - 1

y \;\; = \; \frac{3}{2}

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