# Introduction To Differential Equations

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A differential equation can simply be said to be an equation involving derivatives of an unknown function. For example, consider the equation

\[\frac{{dy}}{{dx}} + xy = {x^2}\]

This is a differential equation since it involves the derivative of the function .\(y(x)\). which we may wish to determine. We must first understand why and how differential equations arise and why we need them at all. In general, we can say that a differential equation describes the behaviour of some continuously varying quantity.

**Scenario - 1 : A freely falling body**

A body is released at rest from a height *h*. How do we described the motion of this body ?

The height *x* of the body is a function of time. Since the acceleration of the body is *g*, we have

\[\frac{{{d^2}x}}{{d{t^2}}} = - g\]

This is the differential equation describing the motion of the body. Along with the initial condition \(x(0) = h,\) it completely describes the motion of the body at all instants after the body starts falling.

**Scenario - 2: Radioactive disintegration**

Experimental evidence shows that the rate of decay of any radioactive substance is proportional to the amount of the substance present, i.e.,

\[\frac{{dm}}{{dt}} = - \lambda m\]

where *m* is the mass of the radioactive substance and is a function of *t*. If we know *m *(0), the initial mass, we can use this differential equation to determine the mass of the substance remaining at any later time instant.

**Scenario - 3 Population growth**

The growth of population (of say, a biological culture) in a closed environment is dependent on the birth and death rates. The birth rate will contribute to increasing the population while the death rate will contribute to its decrease. It has been found that for low populations, the birth rate is the dominant influence in population growth and the growth rate is linearly dependent on the current population. For high populations, there is a competition among the population for the limited resources available, and thus the death rate becomes dominant. Also, the death rate shows a quadratic dependence on the current population.

Thus, if \(N(t)\) represents the population at time *t*, the differential equation describing the population variation is of the form

\[\frac{{dN}}{{dt}} = {\lambda _1}N - {\lambda _2}{N^2}\]

where \({\lambda _1}\) and \({\lambda _2}\) are constants.

Along with the initial population *N *(0), this equation can tell us the population at any later time instant.

These three examples should be sufficient for you to realise why and how differential equations arise and why they are important.

In all the three equations mentioned above, there is only independent variable (the time *t* in all the three cases). Such equations are termed **ordinary differential equations**. We might have equations involving more than one independent variable:

\[~\frac{\partial f}{\partial x}+x\frac{\partial f}{\partial y}={{x}^{2}}\]

where the notation \(\begin{align}\frac{\partial }{{\partial x}}\end{align}\) stands for the partial derivative, i.e., the term \(\begin{align}\frac{\partial }{{\partial x}}\end{align}\) would imply that we differentiate the function *f* with respect to the independent variable *x* as the variable (while treating the other independent variable *y* as a constant). A similar interpretation can be attached to \(\begin{align}\frac{\partial }{{\partial y}}\end{align}\).

Such equations are termed **partial differential equations** but we’ll not be concerned with them in this chapter.

Consider the ordinary differential equation

\[\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}} + {x^2} = c\]

The order of the highest derivative present in this equation is two; thus, we’ll call it a second order differential equation (*DE*, for convenience).

\[\boxed{{\text{The order of a DE is the order of the highest derivative that occurs in the equation}}}\]

Again, consider the DE

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

The degree of the highest order derivative in this DE is two, so this is a DE of degree two (and order three).

\[\boxed{{{\text{The degree of a DE is the degree of the highest order derivative that occurs in the equation,}}\\ \text{when all the derivatives in the equation are made of free of fractional powers.}}{\text{.}}}\]

For example, the DE

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

is not of degree two. When we make this equation free of fractional powers, by the following rearrangement,

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

we see that the degree of the highest order derivative will become four. Thus, this is a DE of degree four (and order two).

Finally, an *n*^{th} **linear DE** (degree one) is an equation of the form

\[{a_0}\frac{{{d^n}y}}{{d{x^n}}} + {a_1}\frac{{{d^{n - 1}}y}}{{d{x^{n - 1}}}} + ... + {a_{n - 1}}\frac{{dy}}{{dx}} + {a_n}y = b\]

where the \({a_i}^\prime s\) and *b* are functions of *x*.

Solving an *n*^{th} order DE to evaluate the unknown function will essentially consist of doing *n* integrations on the DE. Each integration step will introduce an arbitrary constant. Thus, you can expect in general that the **solution of an n**

^{th}order DE will contain*n*independent arbitrary constants.By *n* independent constants, we mean to say that the most general solution of the DE cannot be expressed in fewer than *n *constants. As an example, the second order DE

\[\frac{{{d^2}y}}{{d{x^2}}} + y = 0\]

has its most general solution of the form

\[y = A\cos x + B\sin x.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots (1)\]

(verify that this is a solution by explicit substitution).

Thus, two arbitrary and independent constants must be included in the general solution. We cannot reduce (1) to a relation containing only one arbitrary constant. On the other hand, it can be verified that the function

\[\begin{align}y = a{e^{x + b}}\end{align}\]

is a solution to the second-order DE

\[\frac{{{d^2}y}}{{d{x^2}}} = y\]

but even through it (seems to) contain two arbitrary constants, it is not the general solution to this DE. This is because it can be reduced to a relation involving only one arbitrary constant:

\[\begin{align}y &= a{e^{x + b}}\\& = a{e^x} \cdot {e^b}\\&= c{e^x}\,\,\,\,\,\,\,\,\,\,\,(where\; c = a \cdot {e^b})\end{align}\]

Let us summaries what we’ve seen till now : the most general solution of an *n*^{th} order DE will consist of *n* arbitrary constants; conversely, from a functional relation involving *n* arbitrary constants, an *n*^{th} order DE can be generated (we’ll soon see how to do this). We are generally interested in solutions of the DE satisfying some particular constraints (say, some initial values).Since the most general solution of the DE involves *n* arbitrary constant, we see that the maximum member of independent conditions which can be imposed on a solution of the DE is *n*.

As a first example, consider the functional relation

\[y = {x^2} + {c_1}{e^{2x}} + {c_2}{e^{3x}}\qquad\qquad \ldots (1)\]

This curve’s equation contains two arbitrary constants; as we vary \({c_1}\) and \({c_2},\) we obtain different curves; those curves constitute a family of curves. All members of this family will satisfy the DE that we can generate from this general relation; this DE will be second order since the relation contains two arbitrary constants.

We now see how to generate the DE. Differentiate the given relation twice to obtain

\[\begin{align}& {y}'=2x+2{{c}_{1}}{{e}^{2x}}+3{{c}_{2}}{{e}^{3x}}\qquad\qquad \ldots (2) \\ & y''=2+4{{c}_{1}}{{e}^{2x}}+9{{c}_{2}}{{e}^{3x}} \qquad \qquad \;\,\ldots (3) \end{align}\]

From (1), (2) and (3), \({c_1}\) and \({c_2}\) can be eliminated to obtain

\[\begin{align}& \qquad \;\;\left| {\begin{array}{*{20}{c}}{{e^{2x}}}&{{e^{3x}}}&{{x^2} - y}\\{2{e^{2x}}}&{3{e^{3x}}}&{2x - y'}\\{4{e^{2x}}}&{9{e^{3x}}}&{2 - y''}\end{array}} \right| = 0\\&\Rightarrow \quad \left| {\begin{array}{*{20}{c}}1&1&{{x^2} - y}\\2&3&{2x - y'}\\4&9&{2 - y''}\end{array}} \right| = 0\\& \Rightarrow \quad 6 - 3y'' - 18x + 9y' + 8x - 4y' - 4 + 2y'' + 6{x^2} - 6y = 0\\&\Rightarrow \quad y'' - 5y' + 6y = 6{x^2} - 10x + 2 \qquad \qquad \qquad \qquad \qquad \qquad \ldots (4)\end{align}\]

This is the required DE; it corresponds to the family of curves given by (1). Differently put, the most general solution of this DE is given by (1).

As an exercise for the reader, show that the DE corresponding to the general equation

\[y = A{e^{2x}} + B{e^x} + C\]

where *A*, *B*, *C* are arbitrary constants, is

\[y''' - 3y'' + 2y' = 0\]

As expected, the three arbitrary constants cause the DE to be third order.

- Live one on one classroom and doubt clearing
- Practice worksheets in and after class for conceptual clarity
- Personalized curriculum to keep up with school