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HCU MCA 2013 Paper

Question : 65 of 75

Marks: +1, -0

The solution to the differential equation (1 + xy) ydx + (1 - yx) xdy = 0 is

Solution:

(a) Given differential equation,

( 1 + x y ) y d x + ( 1 - x y ) x d y = 0

\Rightarrow ( y d x + x d y ) + x y^{2} d x - x^{2} y d y = 0

\Rightarrow d ( x y ) = x^{2} y d y - x y^{2} d x

\Rightarrow \frac{d(x y)}{x y} = ( x d y - y d x )

\Rightarrow \frac{d(x y)}{x^{2} y^{2}} = \frac{( x d y - y d x )}{x y} \times \frac{x}{x}

\Rightarrow \frac{d(x y)}{(x y)^{2}} = \frac{x}{y} \cdot \frac{( x d y - y d x )}{x^{2}}

\Rightarrow \frac{d(x y)}{(x y)^{2}} = d \log \left( \frac{y}{x} \right)

On integrating both sides, we get

\int \frac{d(x y)}{(x y)^{2}} = \int d \left( \log \frac{y}{x} \right)

\Rightarrow -\frac{1}{x y} = C + \log \frac{y}{x} = \log \left( \frac{x}{y} \right)^{-1} + C

\Rightarrow -\frac{1}{x y} = - \log \frac{x}{y} + C

\Rightarrow \log(x/y) - \frac{1}{x y} = C

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