Unit - 5

First order ordinary differential equations

A differential equation of the form

Is called linear differential equation.

It is also called Leibnitz’s linear equation.

Here P and Q are the function of x

Working rule

(1) Convert the equation to the standard form

(2) Find the integrating factor.

(3) Then the solution will be y (I.F) =

Example-1: Solve-

Sol. We can write the given equation as-

So that-

I.F. =

The solution of equation (1) will be-

Or

Or

Or

Example-2: Solve-

Sol.

We can write the equation as-

We see that it is a Leibnitz’s equation in x-

So that-

Therefore, the solution of equation (1) will be-

Or

Example-3: Solve sin x )

Solution: here we have,

sin x )

which is the linear form,

Now,

Put tan so that sec² dx = dt, we get

Which is the required solution.

Bernoulli’s equation-

The equation

Is reducible to the Leibnitz’s linear equation and is usually called Bernoulli’s equation.

Working procedure to solve the Bernoulli’s linear equation-

Divide both sides of the equation –

By, so that

Put so that

Then equation (1) becomes-

)

Here we see that it is a Leibnitz’s linear equations which can be solved easily.

Example: Solve

Sol.

We can write the equation as-

On dividing by , we get-

Put so that

Equation (1) becomes,

Here,

Therefore, the solution is-

Or

Now put

Integrate by parts-

Or

Example: Solve

Sol. Here given,

Now let z = sec y, so that dz/dx = sec y tan y dy/dx

Then the equation becomes-

Here,

Then the solution will be-

Example: Solve-

Sol. Here given-

We can re-write this as-

Which is a linear differential equation-

The solution will be-

Put

Exact Differential equations & Reducible to exact

Definition-

An exact differential equation is formed by differentiating its solution directly without any other process,

Is called an exact differential equation if it satisfies the following condition-

Here

is the differential co-efficient of M with respect to y keeping x constant and

is the differential co-efficient of N with respect to x keeping y constant.

Step by step method to solve an exact differential equation-

1. Integrate M w.r.t. x keeping y constant.

2. Integrate with respect to y, those terms of N which do not contain x.

3. Add the above two results as below-

Example-1: Solve

Sol.

Here M = and N =

Then the equation is exact and its solution is-

Example-2: Solve-

Sol. We can write the equation as below-

Here M = and N =

So that-

The equation is exact and its solution will be-

Or

Example-3: Determine whether the differential function ydx –xdy = 0 is exact or not.

Solution. Here the equation is the form of M (x, y) dx + N (x, y) dy = 0

But we will check for exactness,

These are not equal results, so we can say that the given diff. eq. is not exact.

Equation reducible to exact form-

1. If M dx + N dy = 0 be a homogenous equation in x and y, then 1/ (Mx + Ny) is an integrating factor.

Example: Solve-

Sol.

We can write the given equation as-

Here,

M =

Multiply equation (1) by we get-

This is an exact differential equation-

2. I.F. for an equation of the type

IF the equation Mdx + Ndy = 0 be this form, then 1/ (Mx – Ny) is an integrating factor.

Example: Solve-

Sol.

Here we have-

Now divide by xy, we get-

Multiply (1) by , we get-

Which is an exact differential equation-

3. In the equation M dx + N dy = 0,

(i) If be a function of x only = f(x), then is an integrating factor.

(ii) If be a function of y only = F(x), then is an integrating factor.

Example: Solve-

Sol.

Here given,

M = 2y and N = 2x log x - xy

Then-

Here,

Then,

Now multiplying equation (1) by 1/x, we get-

4. For the following type of equation-

An I.F. is

Where-

Example: Solve-

Sol.

We can write the equation as below-

Now comparing with-

We get-

a = b = 1, m = n = 1, a’ = b’ = 2, m’ = 2, n’ = -1

I.F. =

Where-

On solving we get-

h = k = -3

Multiply the equation by , we get-

It is an exact equation.

So that the solution is-

Where, are constant is called homogenous equation.

Put,

Example: Solve

Ans. Put,

AE is

Example:

Ans. Putting,

AE is

CS = CF + PI

Example: Solve

Ans. Let,

AE is

y= CF + PI

Example: Solve

Ans. Let, so that z = log x

AE is

Equation solvable for p-

Example- Solve-

Sol. Here we have-

or

On integrating, we get-

Equation solvable for y-

Steps-

First- differentiate the given equation w.r.t. x.

Second- Eliminate p from the given equation, then the eliminant is the required solution.

Example: Solve

Sol.

Here we have-

Now differentiate it with respect to x, we get-

Or

This is the Leibnitz’s linear equation in x and p, here

Then the solution of (2) is-

Or

Or

Put this value of x in (1), we get

Equation solvable for x-

Example: Solve-

Sol. Here we have-

On solving for x, it becomes-

Differentiating w.r.t. y, we get-

or

On solving it becomes

Which gives-

Or

On integrating

Thus, eliminating from the given equation and (1), we get

Which is the required solution.

References:

1. G.B. Thomas and R.L. Finney, “Calculus and Analytic Geometry”, Pearson, 2002.

2. T. Veerarajan, “Engineering Mathematics”, McGraw-Hill, New Delhi, 2008.

3. B. V. Ramana, “Higher Engineering Mathematics”, McGraw Hill, New Delhi, 2010.

4. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.

5. B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.