Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS4 Marks
Question
Form the differential equation of all circles which pass through origin and whose centres lie on Y-axis.
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Answer
It is given that, circles pass through origin and their centres lie on Y-axis. Let (0, k) be the centre of the circle and radius is k. So, the equation of circle is $(\text{x}-0)^2+(\text{y}-\text{k})^2=\text{k}^2$ $\Rightarrow\text{x}^2+(\text{y}-\text{k})^2=\text{k}^2$ $\Rightarrow\text{x}^2+\text{y}^2-2\text{ky}=0$ $\Rightarrow\frac{\text{x}^2+\text{y}^2}{2\text{y}}=\text{k}\ ......(\text{i})$ On differentiating Eq. (i) w.r.t.x, we get $\frac{2\text{y}\Big(2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)-({\text{x}^2+\text{y}^2})\frac{2\text{dy}}{\text{dx}}}{4\text{y}^2}=0$ $\Rightarrow4\text{y}\Big(\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}\Big)-2(\text{x}^2+\text{y}^2)\frac{\text{dy}}{\text{dx}}=0$ $\Big[4\text{y}^2-2(\text{x}^2+\text{y}^2)\Big]\frac{\text{dy}}{\text{dx}}+4\text{xy}=0$ $\Rightarrow(4\text{y}^2-2\text{x}^2-2\text{y}^2)\frac{\text{dy}}{\text{dx}}+4\text{xy}=0$ $\Rightarrow(2\text{y}^2-2\text{x}^2)\frac{\text{dy}}{\text{dx}}+4\text{xy}=0$ $\Rightarrow(\text{y}^2-\text{x}^2)\frac{\text{dy}}{\text{dx}}+2\text{xy}=0$ $\Rightarrow(\text{x}^2-\text{y}^2)\frac{\text{dy}}{\text{dx}}-2\text{xy}=0$
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