First slide

Formation of differential equations

Question

The differential equation of all circles which passes through the origin and whose centre lies on y-axis is

Easy

A

${(x}^{2} {-y}^{2}) \frac{dy}{dx} - 2xy = 0$
B

${(x}^{2} {-y}^{2}) \frac{dy}{dx} + 2xy = 0$
C

${(x}^{2} {-y}^{2}) \frac{dy}{dx} -xy = 0$
D

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

Solution

$Required circle equation is x^{2} + (y - a)^{2} = a^{2}$
$\begin{array}{l} \to x^{2} + y^{2} - 2 a y = 0 . . . . . . (1) \\ Dwr to x \\ \to 2 x + 2 {yy}^{1} - 2 {ay}^{1} = 0 \\ \to \frac{x + y y^{1}}{y^{1}} = a \dots (2) \end{array}$
$\begin{array}{l} From 1 and 2 we get \to x^{2} + y^{2} - 2 y (\frac{x + {yy}^{1}}{y^{1}}) = 0 \\ \to (x^{2} - y^{2}) \frac{dy}{dx} - 2 xy = 0 \end{array}$

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