Name: Formation of Differential Equations
Uploaded: 2019-07-19
Duration: 11 min 9 s
Description: In this video, we are going to learn about the formation of differential equations in detail. Master this concept in just a few minutes with this quick lesson.

Question

Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes

Accepted Answer

Given circles touches co-ordinates axes lying in the second quadrant, so let centre be $(- a, a)$ then radius is $a$ .

Hence, the equation of the family of circles is $(x - (- a))^{2} + (y - a)^{2} = a^{2}$ , where $a$ is an arbitrary constant.

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

Differentiating the above equation with respect to $x$ , on both sides, we get

$2 (x + a) + 2 (y - a) \frac{d y}{d x} = 0$

$\Rightarrow (x + a) + (y - a) \frac{d y}{d x} = 0$

$\Rightarrow x + a - a \frac{d y}{d x} + y \frac{d y}{d x} = 0$

$\Rightarrow a = \frac{x + y \frac{d y}{d x}}{\frac{d y}{d x} - 1}$ , let $\frac{d y}{d x} = p$

$\Rightarrow a = \frac{x + y p}{p - 1}$

Substituting the value of $a$ in equation (1), we get

${(x + \frac{x + y p}{p - 1})}^{2} + {(y - \frac{x + y p}{p - 1})}^{2} = {(\frac{x + y p}{p - 1})}^{2}$

$\Rightarrow {(\frac{x p - x + x + y p}{p - 1})}^{2} + {(\frac{y p - y - x - y p}{p - 1})}^{2} = {(\frac{x + y p}{p - 1})}^{2}$

$\Rightarrow {(x p + y p)}^{2} + {(y p - y - x - y p)}^{2} = {(x + y p)}^{2}$

$\Rightarrow p^{2} (x + y)^{2} + (y + x)^{2} = {(x + y p)}^{2}$

$\Rightarrow (x + y)^{2} (p^{2} + 1) = {(x + y p)}^{2}$

This is the required differential equation, where $p = \frac{d y}{d x}$ .

Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes

Important Questions on Differential Equations and Their Formation

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