Differential Equations I

  • Differential Equation

An equation consisting of one (or more) independent variables and one (or more) dependent variables and at least one derivative is known as a differential equation (D.E.).

The derivatives can be ordinary or partial. Depending on them, a D.E. can be ordinary D.E. or partial D.E.

[Note the development so far: Set Theory, Relations, Functions, Limit of a Function, Derivatives, Integration, Differential Equations]

  • Order and Degree

Order of a D.E. is the order of the highest ordered derivative in the D.E. Degree of a D.E. is the power to which the highest ordered derivative is raised, provided that D.E. is free from radical and no derivative is present in the denominator.

In this section, we will deal with D.E. of 1 independent variable, x or t, and 1 dependent variable y.

  • Solution of a Differential Equation

The relation y=f(x), which satisfies the given D.E. is known as the solution of that D.E. For example, y= A cos (\omega t) + B sin (\omega t) is a solution of

{\frac {d^2 y}{dt^2} + \omega^2 y = 0}

A and B are arbitrary constants. The solution containing arbitrary constants is known as a general solution. If values of arbitrary constants are given, it becomes a particular solution.

  • Formation of a D.E.

It can be seen that the degree of differential equation is equal to the number of arbitrary constants in its solution. So, to form a D.E. from its solution with n arbitrary constants, we need to differentiate it n times. The arbitrary constants are then to be eliminated.

  • Types of Ordinary D.E. of 1st order and 1st degree
  • Variable-Separable Form

The variables can be integrated separately. D.E. can be written in the form

\int f(x)dx+ \int g(y)dy = C

Illustration : Solve the D.E. {xy \frac {dy}{dx} = (1-x^2)(1+y^2)} (June 2015)

Solution : The terms containing {y} and those containing {x} can be separated. Hence, this is a variable-separable type D.E.

{\frac {y}{1+y^2}dy = \frac {(1-x^2)}{x}dx}


{\frac 1 2 \int \frac {2y}{1+y^2}dy = \int \left (\frac {1}{x} -x \right) dx}

{\frac 1 2 ln (1+y^2) = ln (x) - \frac {x^2}{2} + C}

This is the required solution.

  • Homogeneous Equations

If the sum of powers of x and y in each term of a D.E. is same, it is said to be homogeneous. A homogeneous D.E. can be solved by substituting

y =vx, \frac{dy}{dx} = v + x \frac {dv}{dx}

Illustration : Solve the D.E. {(x^4-2xy^3)dx + (x^4-2xy^3)dy = 0}

  • Non-homogeneous Equations

These are of the form:

\frac {dy}{dx} = \frac {a_1 x + b_1 y + c_1}{a_2x + b_2y + c_2}

Type 1

When \frac {a_1}{a_2} = \frac {b_1}{b_2} = k, write a_1 = ka_2, b_1 = kb_2.

It then reduces to \frac {dy}{dx} = \frac {k (a_2x+b_2y) + c_1}{a_2x + b_2y + c_2}

Writing a_2x+b_2y as v, we get a variable-separable form in x and v.

Type 2

When \frac {a_1}{a_2} \ne \frac {b_1}{b_2} , write x = X + \alpha, y = Y + \beta.

Find \alpha and \beta such that a_1 \alpha + b_1 \beta + c_1 = 0  and a_2 \alpha + b_2 \beta + c_2 = 0

The equation then becomes

\frac {dY}{dX} = \frac {a_1X+b_1Y}{a_2X+b_2 Y}

This is a homogeneous D.E. and can be solved by substitution Y=VX.

  • Exact (or Total) Differential Equations

An equation is said to be exact, when the expression containing differentials dx and dy can be reduced to a single differential du. The solution of this D.E. will then be \int du.

For M dx + N dy to be exact, check for \frac {\partial M}{\partial y} = \frac {\partial N}{\partial y}

The equation can then be solved by:

\int \limits_{y = constant} M dx + \int \limits_{no \ x} N dy = c


\int \limits_{x = constant} N dy + \int \limits_{no \ y} M dx = c

Illustration : Solve the D.E. {\frac {dy}{dx}= \frac {2x-3y+1}{3x+4y-5}} (June 2016)

Solution : Cross-multiplying,

{(2x-3y+1)dx = (3x+4y-5)dy}

{(2x-3y+1)dx + (-3x-4y+5)dy= 0}

Comparing this with {Mdx + Ndy =0}, we get {M = 2x-3y+1} and {N = -3x-4y+5}.

{\frac {\partial M}{\partial y} = -3 \ and \ \frac {\partial N}{\partial x} = -3}

Since {\frac {\partial M}{\partial y}} and {\frac {\partial N}{\partial x}} are equal, this D.E. is exact. The solution is

{\int \limits_{y= constant} M dx + \int \limits_{x \ free} N dy = C}

This gives

{\int \limits_{y= constant} (2x-3y+1)dx + \int \limits_{x \ free} (3x+4y-5)dy = C}

On integrating,

{(x^2 - 3xy + x) + (2y^2-5y) = C}


{x^2-3xy+2y^2+x-5y + C_1 = 0}

Note that this is a general second degree equation, which may represent various conics, depending on the value of {C_1}.

Tip for MCQs : Any differential equation of the form \frac {dy}{dx} = \frac {a_1 x + b_1 y + c_1}{a_2x + b_2y + c_2} will be exact if {b_1+a_2=0}.

  • D.E.s Reducible to Exact D.E.

An integrating factor is the term, with which if a D.E. is multiplied, it becomes exact. Since it is easy to solve an exact D.E., more attention is given to develop methods to find the integrating factor.

If the equation is homogeneous and y \times f(xy)dx + x \times g(xy)dy = 0, then integrating factor is  \frac {1}{Mx + Ny}.

If the equation is of the form y \times f(xy)dx + x \times g(xy)dy = 0, then the integrating factor is \frac {1}{xy[f(xy)-g(xy)]}.

If, for the given D.E.\frac {\frac {\partial M}{\partial y}- \frac {\partial N}{\partial x}}{N} = f(x), then the integrating factor is $e^{\int f(x)dx}$.

If, for the given D.E. \frac {\frac {\partial N}{\partial x}- \frac {\partial M}{\partial y}}{M} = f(y) & $ e^{\int f(y)dy}$

If the given D.E. is of the form x^a y^b (mydx+ nxdy) + x^{\alpha}y^{\beta}(\mu y dx + \nu x dy)=0, then x^{\gamma}y^{\delta}. In this type, we obtain latex \gamma$ and \delta by testing the condition of exactness.

  • Linear D.E.

When the order of a D.E. is 1, it is linear.

The standard form is

{\frac {dy}{dx}+Py = Q}

The solution is given by

{y e^{\int P dx}= \int Q e^{\int P dy}dy}

P and Q are functions of x only.

Video Link (with numericals)

  • Bernoulli’s Equation

This equation is of the form

{\frac {dy}{dx}+ Py = Q y^n}

It can be reduced to linear form by dividing it by y^{n} and then substituting y^{1-n} as u. The equation then becomes a linear equation in u and x.


One thought on “Differential Equations I”

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s