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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Form the differential equation representing the family $y^2-2ay+x^2=a^2$ where $a$ is an arbitrary constant.

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Toolbox:
  • When the function occurs in the form of their products, we can differentiate them using product rule, which states that $uv = uv' + vu' $
Step 1:
The given equation is $y^2-2ay+x^2=a^2$--------(1)
Differentiating with respect to $x$ we get,
$2yy'-2ay'+2x=0$
$\Rightarrow yy'-ay'+x=0$
$\therefore \large\frac{yy'+x}{y'}$$=a$
Step 2:
Now substitute for $a$ in equation(1)
$y^2-2\big(\large\frac{yy'+x}{y'}\big)$$+x^2=\big(\large\frac{yy'+x}{y'}\big)^2$
$\Rightarrow x^2+y^2=\big(\large\frac{yy'+x}{y'}\big)^2$$+2\big(\large\frac{yy'+x}{y'}\big)$
$\Rightarrow x^2+y^2=\big(\large\frac{yy'+x}{y'}\big)\bigg[\large\frac{yy'+x}{y'}$$+2\bigg]$
$\Rightarrow x^2+y^2=\big(\large\frac{yy'+x}{y'}\big)\bigg[\large\frac{yy'+x+2y'}{y'}\bigg]$
$(y')^2[x^2+y^2]-(yy'+x)(yy'+x+2y')=0$
$\therefore x(y')^2-2y(y')^2-2xy'-x^2=0$
This is the required equation.
answered Sep 20, 2013 by sreemathi.v
 

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