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KEAM 2015 Math Paper

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Question : 19 of 120

Marks: +1, -0

The solution of the differential equation

y'(y^2-x)=y

is

$y^3-3xy=C$
$y^3+3xy=C$
$x^3-3xy=C$
$y^3-xy=C$
$x^3-xy=C$

Solution:

y'(y^{2}-x)=y

\Rightarrow \frac{dy}{dx}=\frac{y}{y^{2}-x}

\Rightarrow \frac{dx}{dy}=\frac{y^{2}-x}{y}

\Rightarrow \frac{dx}{dy}=y-\frac{x}{y}

\Rightarrow \frac{dx}{dy}+\frac{1}{y}x=y

Above differential equation is a linear differential equation in

x

.

\text{IF}=e^{\int \frac{1}{y} \omega}=e^{\log y}=y

Hence, solution will be

x \cdot y = \int y \cdot y \, dy + C'

\Rightarrow xy=\frac{y^{3}}{3}+C'

\Rightarrow 3xy=y^{3}+3C'

\Rightarrow y^{3}-3xy=-3C'

\Rightarrow y^{3}-3xy=C

[put

C=-3C',

constant

]

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