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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

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 x-axis.
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Answer

The equation of the family of circles that pass through the origin (0, 0) and whose centres lie on the x-axis is given by

$(\text{x}-\text{a})^2+\text{y}^2=\text{a}^2\ ...(1)$

where a are arbitrary constants.

As this equation has only one arbitrary constant, we shall get a first order differential equation.

Differentiating (1) with respect to x, we get

$2(\text{x}-\text{a})+2\text{y}\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\text{x}-\text{a}+\text{y}\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=\text{a}$

Substituting the value of a in equation (2), we get

$\Big(\text{x}-\text{x}-\text{y}​​\frac{\text{dy}}{\text{dx}}\Big)^2+\text{y}^2=\Big(\text{x}+\text{y}​​\frac{\text{dy}}{\text{dx}}\Big)^2$

$\Rightarrow\text{y}^2\Big(​​\frac{\text{dy}}{\text{dx}}\Big)^2+\text{y}^2=\text{x}^2+2\text{xy}​​\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big(​​\frac{\text{dy}}{\text{dx}}\Big)^2$

$\Rightarrow2\text{xy}​​\frac{\text{dy}}{\text{dx}}+\text{x}^2=\text{y}^2$

It is the required differential equation.

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