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Integral Calculus

Cool goIITian

Joined:	24 Apr 2007
Post:	88

16 Mar 2008 23:00:07 IST

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523

solve the differial equation

Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Integral Calculus

xdy-ydx=[x^2-y^2] dx

Comments (3)

boredom

Blazing goIITian

Joined: 14 Jan 2007

Posts: 602

16 Mar 2008 23:03:51 IST

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HEY,

PUT Y=XT

X(T+ X DT/DX)=XROOT(1-T^2)

kasirajan

Blazing goIITian

Joined: 20 Jan 2007

Posts: 472

16 Mar 2008 23:50:18 IST

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xdy - ydx = ^{[ ]}

x^2 + y^2 dx

xdy/dx = y +

x^2 + y^2 dx

dy/dx = y/x +

1 + { y/x } ^2

put y = vx

y = v + xdv/dx

v + xdv/dx = v +

1 + v^2

1/

1+ v^2 dv = (1/x) dx

on itegratin

log(v +

1 + v^2 ) = log x + C

log ( (y/x) +

1 + (y/x)^2 ) = logx + logA where A > 0

y/x +

(x^2 + y^2 ) /x =Ax

y/x +

(x^2 + y^2 ) = Ax^2.

layman

Hot goIITian

Joined: 15 Mar 2008

Posts: 137

17 Mar 2008 12:47:04 IST

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One very easy method:

$x.dy - y.dx = sqrt {x^{2} - y^{2}} .dx$
=> $rac {x.dy - y.dx}{sqrt {x^{2} - y^{2}}} = dx$
Divide both sides by x,
=> $rac { rac {x.dy - y.dx}{x^{2}}}{sqrt {1 - {( rac {y}{x})}^{2}}} = rac {dx}{x}$
Integrating we get,
=> $sin^{ - 1} { rac {y}{x}} = ln{|x|} + C$

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