Question

Represent the following families of curves by forming the corresponding differential equations (a, b being parameters):
(i) x² + y² = a²

(ii) x² − y² = a²

(iii) y² = 4ax

(iv) x² + (y − b)² = 1

(v) (x − a)² − y² = 1

(vi) $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

(vii) y² = 4a (x − b)

(viii) y = ax³

(ix) x² + y² = ax³

(x) y = e^ax

Answer 1

(i) The equation of the family of curves is
$x^2+y^2=a^2$ ...(1)
where $a$ is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to $x$, we get
$$\begin{gathered} 2x+2y\frac{dy}{dx}=0 \\ \Rightarrow x+y\frac{dy}{dx}=0 \\ \text{It is the required differential equation.} \end{gathered}$$

(ii) The equation of family of curves is
$x^2-y^2=a^2$ ...(1)
where $a$ is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to $x$, we get
$$\begin{gathered} 2x-2y\frac{dy}{dx}=0 \\ \Rightarrow x-y\frac{dy}{dx}=0 \\ \text{It is the required differential equation.} \end{gathered}$$

(iii) The equation of family of curves is
$y^2=4ax$ ...(1)
where $a$ is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to $x$, we get
$$\begin{gathered} 2y\frac{dy}{dx}=4a \\ \Rightarrow 2y\frac{dy}{dx}=\frac{y^2}{x}\left[\text{Using}\left(1\right)\right] \\ \Rightarrow 2x\frac{dy}{dx}=y \\ \Rightarrow y-2x\frac{dy}{dx}=0 \\ \text{It is the required differential equation.} \end{gathered}$$

(iv) The equation of family of curves is
$x^2+{\left(y-b\right)}^2=1$ ...(1)
where $b$ is a parameter.
As this equation contains only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to $x$, we get
$$\begin{gathered} 2x+2\left(y-b\right)\frac{dy}{dx}=0 \\ \Rightarrow 2x+2\sqrt{1-x^2}\frac{dy}{dx}=0\left[\text{Using}\left(\text{1}\right)\right] \\ \Rightarrow x=-\sqrt{1-x^2}\frac{dy}{dx} \\ \Rightarrow x^2=\left(1-x^2\right){\left(\frac{dy}{dx}\right)}^2 \\ \Rightarrow x^2={\left(\frac{dy}{dx}\right)}^2-x^2{\left(\frac{dy}{dx}\right)}^2 \\ \Rightarrow x^2\left[1+{\left(\frac{dy}{dx}\right)}^2\right]={\left(\frac{dy}{dx}\right)}^2 \\ \text{It is the required differential equation.} \end{gathered}$$

(v) The equation of family of curves is
${\left(x-a\right)}^2-y^2=1$ ...(1)
where $a$ is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to $x$, we get
$$\begin{gathered} 2\left(x-a\right)-2y\frac{dy}{dx}=0 \\ \Rightarrow \left(x-a\right)-y\frac{dy}{dx}=0 \\ \Rightarrow \sqrt{1+y^2}=y\frac{dy}{dx}\left[\text{Using}\left(\text{1}\right)\right] \\ \Rightarrow 1+y^2=y^2{\left(\frac{dy}{dx}\right)}^2 \\ \Rightarrow y^2{\left(\frac{dy}{dx}\right)}^2-y^2=1 \\ \text{It is the required differential equation.} \end{gathered}$$

(vi) The equation of family of curves is
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ ...(1)
where $a\mathrm{and}b$ are parameters.
As this equation has two arbitrary constants, we shall get a differential equation of second order.
Differentiating (1) with respect to $x$, we get
$\frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0$, ...(2)
Differentiating (2) with respect to $x$, we get
$$\begin{gathered} \frac{2}{a^2}-\frac{2}{b^2}{\left(\frac{dy}{dx}\right)}^2-\frac{2y}{b^2}\frac{d^{2}y}{d{x}^2}=0 \\ \Rightarrow \frac{2}{a^2}=\frac{2}{b^2}\left[y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2\right] \\ \Rightarrow \frac{b^2}{a^2}=\left[y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2\right]...\left(\text{3}\right) \end{gathered}$$
Now, from (2), we get
$$\begin{gathered} \frac{2x}{a^2}=\frac{2y}{b^2}\frac{dy}{dx} \\ \Rightarrow \frac{b^2}{a^2}=\frac{y}{x}\frac{dy}{dx}...\left(4\right) \end{gathered}$$
From (3) and (4), we get
$$\begin{gathered} \frac{y}{x}\frac{dy}{dx}=\left[y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2\right] \\ \Rightarrow x\left[y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2\right]=y\frac{dy}{dx} \\ \text{It is the required differential equation.} \end{gathered}$$

(vii) The equation of family of curves is
$y^2=4a\left(x-b\right)$ ...(1)
where $a\mathrm{and}b$ are parameters.
As this equation has two arbitrary constants, we shall get a differential equation of second order.
Differentiating (1) with respect to $x$, we get
$$\begin{gathered} 2y\frac{dy}{dx}=4a \\ \Rightarrow y\frac{dy}{dx}=2a...\left(\text{2}\right) \end{gathered}$$
Differentiating (2) with respect to $x$, we get
$y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2=0$

It is the required differential equation.

(viii) The equation of family of curves is
$y=a{x}^3$ ...(1)
where $a$ is a parameter.
As this equation has only one arbitrary constant, so we shall get a differential equation of first order.
Differentiating (1) with respect to $x$, we get
$$\begin{gathered} \frac{dy}{dx}=3a{x}^2 \\ \Rightarrow \frac{dy}{dx}=3\times \frac{y}{x^3}\times x^2\left[\text{Using}\left(\text{1}\right)\right] \\ \Rightarrow x\frac{dy}{dx}=3y \\ \text{It is the required differential equation.} \end{gathered}$$

(ix) The equation of family of curves is
$x^2+y^2=a{x}^3$ ...(1)
where $a$ is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to $x$, we get
$$\begin{gathered} 2x+2y\frac{dy}{dx}=3a{x}^2 \\ \Rightarrow 2x+2y\frac{dy}{dx}=3\left(\frac{x^2+y^2}{x^3}\right)x^2\left[\text{Using}\left(\text{1}\right)\right] \\ \Rightarrow 2x+2y\frac{dy}{dx}=3\frac{x^2+y^2}{x} \\ \Rightarrow 2x^2+2xy\frac{dy}{dx}=3x^2+3y^2 \\ \Rightarrow 2xy\frac{dy}{dx}=x^2+3y^2 \\ \text{It is the required differential equation.} \end{gathered}$$

(x) The equation of family of curves is
$$\begin{gathered} y=e^{ax} \\ \Rightarrow \log y=ax...\left(1\right) \end{gathered}$$
where $a$ is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to $x$, we get
$$\begin{gathered} \frac{1}{y}\frac{dy}{dx}=a \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{\log y}{x}\left[\text{Using}\left(\text{1}\right)\right] \\ \Rightarrow x\frac{dy}{dx}=y\log y \\ \text{It is the required differential equation.} \end{gathered}$$