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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Form the differential equation representing the family of curves given by $(x-a)^2+2y^2\;=\;a^2$, where $a$ is an arbitrary constant

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Toolbox:
  • If the given family of curves depends only on one parameter then its is represented by an equation of the form $F(x,y,a) = 0$
  • Differentiating this equation with respect to $x$, we get an equation involving $y'$,$y$ and $x$ and a .i.e; $g(x,y,y',a) = 0 $
  • The required differential equation is then obtained by eliminating a from the above equations.
Step 1:
From the given information we get the equation as
$(x-a)^2 + 2y^2 = a^2$------- (1)
Using the information in the tool box,let us differentiate this equation with respect to $x$ on both sides,
$2(x-a) + 4y.y' = 0$
Step 2:
Dividing throughout by 2 we get,
$x-a +4yy' = 0$
$x + 4yy' = a$---------(2)
substituting in equ(1) we get
$([x -(x+4yy')^2 + 2y^2 = (x+4yy')^2 $
Step 3:
On expanding we get,
$x^2 + (x+4yy')^2 -2x(x+yy') +2y^2 =( x + 4yy')^2$
On simplifying we get,
$x^2 -2x^2 -4xyy' +2y^2 = 0$
$2y^2 - x^2 = 4xyy'$
$\large\frac{[2y^2 - x^2]}{4xy}$$=y'$
This is the required equation.
answered Aug 16, 2013 by sreemathi.v
 

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