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CGPET 2015 Solved Paper

Section: Mathematics

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Question : 114 of 150

Marks: +1, -0

The solution of the differential equation

{√{a+x}} \;{d y}/{d x}+x y=0

$y=C e^{\;{2}/{3}(2 a-x) {√{x+a}}}$
$y= {Ce}^{\;{2}/{3}(a-x) {√{x+a}}}$
$y=C e^{\;{2}/{3}(2 a+x) {√{x+a}}}$
$y=C e^{-\;{2}/{3}(2 a-x) {√{x+a}}}$

Solution:

Given differential equation is

√{a+x} {dy}/{dx} +{xy} = 0

⇒

√{a+x} {dy}/{dx} = {-xy}

⇒

{dy}/y = {-x}/√{a+x} {dx}

On integrating both sides, we get

\log y - \log C = \int[{-x-a}/√{a+x} + a/√{a+x}]{dx}

⇒

\log (y/C) = - \int√{x+a} {dx} +a \int1/√{a+x} {dx}

= - {(x+a)^{3/2}}/{3/2}+ a {(a+x)^{1/2}}/{1/2}

⇒

\log (y/C) = - {2/3} (x+a) √{x+a} +2a √{x+a}

={2/3} (2a-x)√{x+a}

⇒

y/C e^{2/3(2a-x)√{x+a}}

⇒

y=C.e^{2/3(2a-x)√{x+a}}

which is the required solution.

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