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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find the equation of a curve passing through the point (0,-2) given that at any point (x,y) on the curve,the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

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  • $\int x = \large\frac{x^2}{2}$$ + C$
Step 1:
According to the given information we get the equation as $y\large\frac{dy}{dx}$$ = x $
$ydy = xdx$
Integrating on both sides we get,
$\large\frac{y^2}{2} =\frac{ x^2}{2}$$ + C$
$y^2 - \large\frac{x^2}{2}$$ = C$
Step 2:
Since the curve passes through (0,-2), we substituting for $x$ and $y$ to find the value for C
$(-2)^2 - 0 = 2C$
$C =\large\frac{ 4}{2}$$ = 2$
Substituting this we get
$y^2 - x^2 = 4.$
This is the required solution.
answered Aug 14, 2013 by sreemathi.v
 

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