Integral Calculus

maga ishwarya's Avatar
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Joined: 24 Apr 2007
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16 Mar 2008 23:00:07 IST
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solve the differial equation
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Integral Calculus

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



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boredom's Avatar

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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's Avatar

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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's Avatar

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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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