Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Form the differential equation of all the circle which pass through the origin and whose centres lies in $y-$axis.
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Answer
We know that, equation of a dide with centre at $(h, k)$ and radius $r$ is given by,
$(x - h^2) + (y - k)^2 = r^2 ...(1)$
Here, entre lies, on y-axis, so $h = 0$
$x^2 + (y - k)^2 = r^2...(2)$
Also given that circle is passing through origin, so
$0 + k^2 = r^2$
$k^2 = r^2$^
So, equation $(2)$ becomes
$x^2 + (y - k)^2 = k^2$
$x^2 + y^2 - 2yk = 0$
$2yk = x^2+ y^2$
$\text{k}=\frac{\text{x}^2+\text{y}^2}{2\text{y}}$
Differentiating with respect to $x,$
$0=\frac{\Big(2\text{y}2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)-(\text{x}^2+\text{y}^2)2\frac{\text{dy}}{\text{dx}}}{(2\text{y})^2}$
$0=4\text{xy}+4\text{y}^2\frac{\text{dy}}{\text{dx}}-2\text{x}^2\frac{\text{dy}}{\text{dx}}-2\text{y}^2\frac{\text{dy}}{\text{dx}}$
$0=2\text{y}^2\frac{\text{dy}}{\text{dx}}-2\text{x}^2\frac{\text{dy}}{\text{dx}}+4\text{xy}$
$\text{x}^2\frac{\text{dy}}{\text{dx}}-\text{y}^2\frac{\text{dy}}{\text{dx}}=2\text{xy}$
$(\text{x}^2-\text{y}^2)\frac{\text{dy}}{\text{dx}}=2\text{xy}$
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