1.1 Exact differential equations | unit 1 first order ordinary differential equations

Unit – 1

First Order Ordinary Differential Equation

1.1 Exact differential equations

Introduction :-

We know that how to form a differential equation from a given primitive. Now in this chapter we find the primitive from a given differential equation.

The process of obtaining such a primitive from a given differential equation is called solving the differential equation.

Differential Equation :-

An equation involving derivatives or differential coefficient is called differential equation.

Moreover if these derivatives are ordinary, then it is called ordinary differential equation and if they are partial then corresponding equation is called partial differential equation.

Order of the differential Equation :-

The order of differential equation is the order of highest order derivative occurring in that equation.

Eg. i). + cosx = 0

is of order 1

ii). – + 6y = 0 is of order 2.

Degree of the differential Equation :-

The degree of the differential eqn is the power / degree / index of the highest order derivative occurring in that equation.

Eg. (i). + sinx = 0

Is of degree 1.

(ii). ()2 – 5y = 0

Is of degree 2.

First order First degree differential equationn :-

An equation of the form,

Where and are functions in & is called first order first degree differential Equation.

Solution of the differential Equation :-

Any relationship between dependent and independent variable, which satisfies the given differential equation is called solution of that differential Equation.

There are different methods for solving the differential equation of first order and first degree differential equation.

Viz.

(1). Exact differential Equation

[reducible to exact differential equation]

(2). Linear differential Equation

[ Reducible to Linear differential Equation ]

(3). Bernoulli’s differential Equation

Exact differential equation :-

A differential equation which is obtained from its primitive by differentiation only & without any operation of elimination or reduction is called exact differential equation.

Thus if u = c, where u is a function of & is a primitive then,

du = +

Rule for finding the solutions :

If is an exact differential equation, it must have been obtained by equating to zero, total differential du of some function u of & .

Since,

du = +

i.e. M = & N = then

we get a part of u & it’s partial derivative w. r .t is in . the term in u which are free from x do not appear in M but they appear in N. Hence the remaining part of u is obtained by integrating w.r.t. y those terms in N. which are free from x. thus we get,

The Rule :-

Integrate w.r.t. x treating y const, Integrate only those terms in N which are free from x w.r.t. y. equate the sum to a constant which is the required solution.

Ex : 1. Solve( tany + 3 ) dx + ( x sec2y – 3y ) dy = 0

Sol)

Here M = tany + x & N = xsec2y – 3y

= sec2y = Given equation is exact.

Now,

= ∫ ( tany + x ) dx

The solution is

Ex.2. solve(secx tanx tany - ex ) dx + secx sec2y dy = 0

Sol)

Here M = secx tanx tany

= secx tanx sec2y

& N = secx sec2y

= secx tanx sec2y

Given equation is an exact differential equation .

Now

Thus

which is the required solution

Ex3) Solve(sinx cosy + e2x)dx + (cosx siny + tany) dy = 0

Sol)

Here M =

& N =

Given equation is an exact differential equation .

Now

The solution is

which is the required solution

2.2 Equations reducible to exact form

Equations Reducible to Exact differential Equation.

Sometimes the given differential equation is not exact but we can convert it into exact differential equation by multiplying it by a suitable factor, called integrating factor.

Then resultant equation can be solved by exact differential Equation method.

Rules for finding Integrating factor (I.F.)

Rule 1:-If is a function in x only or constant say f(x) then is an integrating factor.

Ex1) Solve (x2 + y2 + 1)dx – 2xydy = 0

Sol)

Here M = x2 + y2 + 1, N =

= 2y, = -2y

Given differential equation is not exact.

Now consider,

= = f(x)

I.F. = =

Multiplying IF to given diff equation we get an exact differential equation as,

[ 1 + y2/x2 + 1/x2 ] dx – 2y/x dy = 0

Here

Hence its solution is given by,

i.e

i.e.

is the required solution of given differential equation.

Exercise

Solve the following

(i). (x4ex – 2mxy2)dx + 2mx2y dy = 0

(ii). (2x2 - y)dx + (xy2 + 1)xdy = 0

(iii). (xsec2y – x2cosy)dy = (tany – 3x4)dx

Rule 2:- If is a function of y only or constant say f(y) then is an Integrating factor.

Ex) Solve(xy3 + y)dx + 2(x2y2 + x + y4)dy = 0

Sol)

Here M = xy3 + y, N = 2(x2y2 + x + y4)

= 3xy2 + 1, = 2(2xy2 + 1) = 4xy2 + 2

Given differential equation is not exact .

Now consider,

= =

IF =

Thus multiplying given equation by I.F. we get an exact differential equation as,

(xy4 + y2)dx + 2(x2y3 + xy + y5)dy = 0

Here

Hence it’s solution is given by,

i.e

i.e.

Is the required solution.

Exercise

(i). (x2 + y2 + x)dx + xydy = 0

(ii). xex(dx - dy) + exdx + yeydy = 0

(iii). ydx – xdy = 0

Rule 3 :- If the equation is of the form f(xy) y dx + g(xy)xdy = 0 and

Mx – Ny ≠ 0 then 1is an integrating factor.

Rule 4 :- if the equation is of the form Mdx + Ndy = 0 where M and N are homogeneous expressions of same degree in x & y and Mx + Ny ≠ 0 then is an integrating factor.

Ex) Solve y(xy + 2x2y2)dx + x(xy – x2y2)dy = 0

Sol)

Here

Thus

Given differential equation is not exact.

Now consider,

Then

P.I = =

Now multiplying given equation by I.F. we get an exact differential Equation as

Here,

Hence its solution is given by,

is the required solution of given differential equation.

Exercise :-

Solve the following.

(i). y(1 + xy + x2y2)dx + x(1 – xy + x2y2)dy = 0

(ii). y(sinxy + xy cosxy)dx + x(xy cosxy - sinxy)dy = 0

(iii). (x2y – 2xy2)dx – (x3 – 3x2y)dy = 0

(iv). x2ydx – (x3 + y3)dy = 0

(v). y(x + y)dx – x(y - x)dy = 0

1.3 Linear differential equations

Linear Equations :-

A differential equation is said to be linear if the dependent variable and its derivatives appears only in the first degree. The form of the linear differential equation of the first order is

Where P & Q are functions of x or constant only.

A linear equation of the form is not solvable as it is.

However if we multiply it by the factor it becomes exact & hence can be solved by the usual method.

To solve the equation :-

Multiply the given equation by the I.F. , we get,

Since, it is exact, we have

Since there is no term in N free from x, the solution is

Remark :-To solve a linear differential equation, first write the equation with the coefficient of unity.

i.e. in the form

Then find and further

Then the solution is.

Ex) sin2x = y + tanx

The given equation can be written as,

which is a linear diff. eqn.

Now

Thus the solution is.

Which is the required solution.

Ex) +

Sol)

This is a linear differential equation

Considr

Hence it’s solution is,

Exercise:

Another form of linear differential equation :-

A differential equation of the form

where & are functions of y only is also a linear differential eqn with x & y having interchanged positions & Hence it’s solution is

Ex)

Sol)

The given eqn can be written as,

which is a linear diff eqn of the form

Now

Hence it’s solution is

Thus

Is the required solution of given du=ifferential equation.

Ex) (1 + x + xy2)dy + (y + y3)dx = 0

Sol)

The given eqn can be written as,

which is of the form,

Now,

Hence solution is,

1.4 Equations reducible to linear form

Equation reducible to the linear form.

An equation of the type

where P and Q are functions of x only can be reduced to the lined form as,

put

given eqn becomes

which is a linear & it’s solution is

Simillarly an equation of the form.

can be reduced to

& it’s soln is

Ex)

Sol)

Dividing by we get

Put

Hence it’s solution is

Put

Thus

i.e.

Ex)

Sol)

Dividing by we get

Put

Thus we get,

which is a linear equation

Now consider

Thus

I.F.==

Hence it’s solution is given by ,

Thus

Is the required solution.

Exercise

1.5 Bernoulli’s Equation :-

An equation of the form

where P and Q are functions of x alone, is called Bernoulli’s equation & it is reducible to linear form by dividing it by and then putting v. after simplification it can be solved by previous method.