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Differential Equations Part 1

Section: Mathematics

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Question : 46 of 100

Marks: +1, -0

If

x^{3} dy + xy dx

= x^{2} dy + 2y dx ; y(2)=e

x

>1,

y(4)

[3 Sep 2020 Shift 2]

$\frac{3}{2} + \sqrt{e}$
$\frac{3}{2} \sqrt{e}$
$\frac{1}{2} + \sqrt{e}$
$\frac{\sqrt{e}}{2}$

Solution:

x^{3} dy + xy dx = x^{2} dy + 2y dx

\Rightarrow dy (x^{3} - x^{2}) = dx (2y - xy)

\Rightarrow -\int \frac{1}{y} dy = \int \frac{x-2}{x^{2}(x-1)} dx

\Rightarrow -\ln y = \int \left( \frac{A}{x} + \frac{B}{x^{2}} \right)

+\left(\frac{C}{x-1}\right) dx

Where

A=1, B=+2, C=-1

\Rightarrow -\ln y = \ln x - \frac{2}{x} - \ln (x-1) + \lambda

\Rightarrow y(2)=e

\Rightarrow -1 = \ln 2 - 1 - 0 + \lambda

\therefore \lambda = -\ln 2

\Rightarrow \ln y = -\ln x + \frac{2}{x}

+ \ln (x-1) + \ln 2

Now put x = 4 in equation

\Rightarrow \ln y = -\ln 4 + \frac{1}{2}

+ \ln 3 + \ln 2

\Rightarrow \ln y = \ln \left( \frac{3}{2} \right) + \frac{1}{2} \ln e

\Rightarrow y = \frac{3}{2} \sqrt{e}

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