Lecture 2 [MATH 129]

Differential Equation

Formation of Differential Equation

    Question: Find the differential equation that describes the family of circles passing through the origin.

    Solution: The general form of the equation of the circle is 

  \begin{align*} & (x - h)^2 + ( y - k )^2 = \sqrt{(h^2 + k^2)^2} \\ \implies & x^2 - 2 xh + h^2 + y^2 - 2yk + k^2 = h^2 + k^2\\ \implies & x^2 - 2xh + y^2 - 2ky = 0 \tag{1.1}\label{eq:1.1} \end{align*}  

Using differentiation twice we get,

  \begin{align*} & 2x - 2h + 2yy' - 2ky' = 0 \\ \implies & x - h + yy' - ky'= 0 \tag{1.2}\label{eq:1.2} \\ and~ & 1 - 0 + yy' + y'y' - ky'' = 0 \\ \implies & 1 + yy' + (y')^2 - ky'' = 0 \tag{1.3}\label{eq:1.3} \end{align*}  

From the equation (1.1) we get,

  \begin{align*} 2xh & = x^2 + y^2 - 2ky \\ h & = \frac{(x^2 + y^2 - 2ky)}{2x} \tag{1.4}\label{eq:1.4} \end{align*}  

From equation (1.2) and (1.4) we get,

  \begin{align*} & x - \frac{x^2 + y^2 - 2ky}{2x} + yy' - ky' = 0 \\ \implies & 2x^2 - x^2 - y^2 + 2ky + 2xyy' - 2xky' = 0 \\ \implies & 2k (y - xy') = - x^2 + y^2 - 2xyy' \\ \implies & ~~~~~~~~~~ k = \frac{- x^2 + y^2 - 2xyy'}{2(y - xy')} \\ \implies & ~~~~~~~~~~ k = \frac{x^2 - y^2 + 2xyy'}{2(xy' - y)} \tag{1.5}\label{eq:1.5} \end{align*}  

Again from (1.3) and (1.5) we get,

  \begin{align*} & 1 + yy'' + (y')^2 - \frac{x^2 - y^2 + 2xyy'}{2(xy' - y)} y'' = 0 \\ \implies & 2[ 1 + (y')^2] (xy' - y) + 2yy'' (xy' - y) - (x^2 - y^2 + 2xyy')y'' = 0 \\ \implies & 2[ 1 + (y')^2] (xy' - y) + 2xyy'y'' - 2y^2 y'' - x^2 y'' + y^2 - 2xyy'y'' = 0 \end{align*}  

Which is the required differential equation.

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    Question: Find the differential equation of the family  $$ y = \frac{2ce^{2x}}{1 + c e^{2x}}$$  

    Solution: 

Given Equation, 

  \begin{align*} y & = \frac{2ce^{2x}}{1 + c e^{2x}} \\ \frac{dy}{dx} & = \frac{(1 + ce^{2x})~ \frac{d}{dx}(2ce^{2x})- 2ce^{2x}\frac{d}{dx}(1 + c e^{2x})}{(1 + c e^{2x})^2} \\ & = \frac{(1 + ce^{2x})~4ce^{2x} - 2ce^{2x}~ 2ce^{2x}}{(1 + ce^{2x})^2} \\ & = \frac{4ce^{2x} + 4c^2 e^{2x} ~e^{2x} - 4c^2 e^{2x} ~e^{2x}}{(1 + ce^{2x})^2} \\ & = \frac{4ce^{2x}}{(1 + ce^{2x})^2} \\ & = \left(\frac{2ce^{2x}}{1 + ce^{2x}}\right)^2 ~ \frac{e^{-2x}}{c} \\ \implies \frac{dy}{dx} & = \frac{y^2 e^{-2x}}{c} \tag{2.1}\label{eq:2.1} \end{align*}  

Again, 

  \begin{align*} & ~~~~~~~~~ y = \frac{2ce^{2x}}{1 + c e^{2x}} \\ \implies & y(1 + c e^{2x}) = 2ce^{2x} \\ \implies & y + cye^{2x} = 2ce^{2x} \\ \implies & ce^{2x} (y - 2) = - y \\ \implies & ~~~~~~~ ce^{2x} = \frac{y}{2 - y} \\ \implies & ~~~~~~~~~ c = \frac{ye^{-2x}}{2 - y} \tag{2.2}\label{eq:2.2} \end{align*}  

From (2.1) and (2.2) we get, 

  \begin{align*} & \frac{dy}{dx} = y^2 e^{- 2x} \frac{1}{\frac{e^{- 2x}y}{2 - y}} \\ & \frac{dy}{dx} = y(2 - y) \end{align*}  

Which is the required differential equation.