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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Form the differential equation of the family of hyperbolas having foci on $x$-axis and centre at origin.

$\begin{array}{1 1}(A)xyy'' + x(y')^2 - y' = 0 \\(B)xyy'' - x(y')^2 - yy' = 0 \\(C)xyy'' + x(y')^2 + yy' = 0 \\ (D)xyy'' + x(y')^2 - yy' = 0 \end{array} $

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Toolbox:
  • The equation of a hyperbola having foci on $x$-axis and centre at origin is $\large\frac{x^2}{a^2} -\frac{ y^2}{b^2}$$ = 1$
Step 1:
From the information in the tool box we take the equation of the hyperbola as $\large\frac{x^2}{a^2} -\frac{ y^2}{ b^2 }$$= 1$-----------(1)
Differentiating this on both sides we get,
$\large\frac{2x}{a^2} - [\frac{1}{b^2}]$$[(2y'y'] = 0$
on simplifying we get $\large\frac{x}{a^2} -(\frac{yy'}{b^2})$ = 0------(2)
Step 2:
Again differentiating this we get
$\large\frac{1}{a^2} - (\frac{1}{b^2})$$ [ (y')^2 + yy''] = 0$
$\large\frac{1}{a^2} = [\frac{1}{b^2}]$$[(y')^2 + yy'']$
Substituting this in equ (2) we get
$\large\frac{x}{b^2}$$[(y')^2 + yy''] - \large\frac{yy'}{b^2}$$ = 0$
Step 3:
Multiplying by $b^2$ throughout and simplifying we get
$xyy'' + x(y')^2 - yy' = 0$
This is the required differential equation.
answered Aug 15, 2013 by sreemathi.v
 

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