# Simple Examples On Differential Equations

**Example – 1**

Find the DE corresponding to the family of rectangular hyperbolas \(xy = {c^2}.\)

**Solution: **Since the equation for a rectangular hyperbola contains only one arbitrary constants, the corresponding DE for the family of rectangular hyperbolas will be first order and can be obtained by differentiation once.

\[\begin{array}{l} & xy = {c^2}\\ \Rightarrow & xy' + y = 0\end{array}\]

This is the required DE.

**Example – 2**

Find the DE associated with the family of circles of a fixed radius *r*.

**Solution: **The circles are of a fixed radius but their centres are not. Let the centre be denoted by the variable point \((h,\,k).\)

Then the equation of an arbitrary circle of the family is

\[{(x - h)^2} + {(y - k)^2} = {r^2} \qquad \dots (1)\]

This contains two arbitrary constants and therefore will give rise to a second-order DE.

Differentiating (1), we have

\[(x - h) + (y - k)\frac{{dy}}{{dx}} = 0\, \qquad\qquad \ldots (2)\]

Differentiating (2) again, we have

\[1 + (y - k)\frac{{{d^2}y}}{{d{x^2}}} + {\left( {\frac{{dy}}{{dx}}} \right)^2} = 0 \qquad \qquad \dots(3)\]

\[\begin{align} \Rightarrow \qquad (y - k) = - \frac{{1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}}}{{\frac{{{d^2}y}}{{d{x^2}}}}} \qquad \qquad \dots(4) \end{align}\]

Using (4) in (2), we have

\[\begin{align}(x - h) = \frac{{\left\{ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right\}\frac{{dy}}{{dx}}}}{{\frac{{{d^2}y}}{{d{x^2}}}}}\end{align}\qquad \qquad \dots(5)\]

Using (4) and (5) in (1), and simplifying, we have the required DE as

\[\begin{align}{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^3} = {r^2}{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2}\end{align}\]

which as expected, is second order.

If all this talk about arbitrary constants and solutions to DEs, confuses you, lets view the whole discussion from a slightly different perspective.

Referring to example -1, suppose we are given the DE

\[\begin{align}xy' + y = 0 \qquad \qquad \dots(1) \end{align}\]

This is a first-order DE and its most general solution will contain one arbitrary constant; in fact, the most general solution of this DE is

\[xy = \lambda \]

where \(\lambda\) is an arbitrary constant. Now suppose we are told that the curve satisfying the DE in (1) passes through (2, 2). This additional constraint enables us to determine the value of \(\lambda \) for this particular curve:

\[(2)(2) = \lambda \quad \Rightarrow \quad \lambda = 4\]

and thus the curve \(xy = 4\) is a particular solution of the DE in (1).

To emphasize once more,

\(xy = \lambda \) is a general solution to (1)

while \(xy = 4\) is a particular solution to (1)

which was obtained from the general solution by using the fact that the curve passes through (2, 2)

As another example, consider the DE obtained in example - 2.

\[{\left( {1 + {{\left( {y'} \right)}^2}} \right)^3} = {r^2}{\left( {y''} \right)^2}\]

This is a second-order DE and its most general solution will contain two arbitrary constants; the most general solution can be found to be

\[{\left( {x - \alpha } \right)^2} + {\left( {y - \beta } \right)^2} = {r^2}\]

Where \(\alpha \) and \(\beta \) are arbitrary constants .

To determine a particular solution to this DE, we need **two** additional constraints which can enable us to evaluate \(\alpha \) and \(\beta \).

**Example – 3**

Find the DE associated with the family of straight lines, each of which is at a constant distance *p* from the origin.

**Solution:** Any such line has the equation

\[x\cos \alpha + y\sin \alpha = p \qquad\qquad \dots (1)\]

where \(\alpha \) is a variable. Different values of \(\alpha \) give different lines belonging to this family. Since the equation representing this family contains only one arbitrary constant, its corresponding DE will be first order.

Differentiating (1), we have

\[\begin{align} & \qquad \;\; \cos \alpha + y'\sin \alpha = 0\\&\Rightarrow \quad \tan \alpha = - \frac{1}{{y'}}\\&\Rightarrow \quad \sin \alpha = \frac{{ - 1}}{{\sqrt {1 + {{(y')}^2}} }},\,\,\,\,\,\cos \alpha = \frac{{y'}}{{\sqrt {1 + {{(y')}^2}} }} \qquad \qquad \ldots (2)\end{align}\]

Using (2) in (1), we have

\[\begin{align} &\qquad\;\;\; \frac{{xy'}}{{\sqrt {1 + {{(y')}^2}} }} - \frac{y}{{\sqrt {1 + {{(y')}^2}} }} = p\\ &\Rightarrow \quad {(xy' - y)^2} = {p^2}(1 + {(y')^2})\end{align}\]

As expected, this is a first order DE.

## TRY YOURSELF - 1

**Q. 1** Find the DE of all circles in the first quadrant which touch the coordinate axes.

**Q. 2** Find the DE of all circles touching the *x*-axis at the origin.

**Q. 3** Find the DE of all hyperbolas having their axes along the coordinate axes.