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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Form the differential equation of the family of parabolas having vertex at origin at axis along positive y-axis.

$\begin{array}{1 1} (A)\;y' -2y = 0 \\(B)\;xy' -y = 0 \\ (C)xy' +2y = 0 \\ (D)xy' -2y = 0 \end{array} $

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  • Equation of a parabola having vertex at origin at axis along the positive axis is $x^2 = 4ay $
Step 1:
Given: $ x^2 = 4ay$
Differentiating on both sides we get,
$2x = 4ay'$
dividing by 2 on both sides,
$x = 2ay'$
$a =\large\frac{ x}{2y'}$
Step 2:
Substituting in equ(1) we get
$x^2 = 4. [\large\frac{x}{2y'}] $$.y$
On simplifying and rearranging we get
$xy' -2y = 0$
This is the required equation.
answered Aug 15, 2013 by sreemathi.v

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