UNIT-1

Ordinary differential equations

1.1 Ordinary differential equations & seperable equations

A differential equation is an equation with a function and one and more of its derivatives:

For example:      y + = 8x

A differential equation with only single independent variable is known as ordinary differential equation.

Here we will learn about degree and order of an ordinary differential equation:

Order: the order is the highest derivative:

For example:  (1)    + y³ = 8x ,  here it has only first derivative so its ‘first order’.

(2)  + y³ = 7x ,  here it has a second derivative so its ‘second order’  and so on.

Degree:  Degree is known as the exponent of higher derivative.

For example:  (1)   )² + y³ = 8x, here its highest derivative has an exponent of 2, it is a first order second degree ordinary differential equation.

(2)  ) + y³ = 8x,  here in this example its higher derivative has no exponent, so we can say that this is the third order and first degree ordinary differential equation.

Note-  A solution to a differential equation which contains one or more arbitrary constants of integration is called general solution.

If additional information is given so that constant may be calculated the particular solution . Additional informations are boundary conditions.

Let’s understand how to solve differential equations:

(1) The solution of equations of the form 

This type of equation can be solved by direction integration:

Y = 

Example-1:  Find the general solution of   x2 – 4x³

Solution: rearranging the given equation:

= - =

On integrating both sides,

y = = 2 In x - + c

This is the general solution.

Exampl-2:  Find the particular solution of the differential equation  5, given the boundary condition y =  1 , when x = 2.

Solution: rearrange the diff. Equation,

=  -

y =   = - + c,    which is the general solution.

Put the boundary conditions to find c,

1 -    , which gives, c = 1

Hence the particular solution is,

y = - + 1.

(2) The solution of equation of the form  ,

Example-1:  Find the general solution of   = 3 + 2y

Solution:      here,   = 3 + 2y gives,

,

Integrating both sides,

,

By substitution,  u = (3 +2y),

X =  In(3 + 2y) + c.

Example-2: Determine the particular solution of  (y² - 1)3y  given that y =1  when  x = 2.

Solution:  It gives,

When putting the values, y =1 , x = 2 ,

The particular solution will be,

(3) The solution of equation of the form 

Example-1: Solve the equation 4xy = y² - 1

Solution:  on separating variables, we get

() dy  =  dx

=  

Using substitution, u = y² - 1

2In(y² - 1) = In x + c.

Example-2: Determine the particular solution of =  2 , given that t = 0, when θ = 0

Solution:    =  2  =  2    ,

= 2dt

dθ =  dt

Now integrating both sides,

=

The general solution is ,

+ c.

When t = 0 and θ = 0,    c =   

= 

1.2. Exact differential equation and bernoulli’s equations:

Definition- A differential equation  M(x , y)dx + N(x , y)dy = 0 is said to be exact if there exist a function h( x, y) such that,

d h(x, y) = M(x , y)dx + N(x , y)dy

Note – before solving exact differential equations , first we test for exactness as mentioned below-

Any differential equation is exact if and only if-

=

Working steps to solve exact differential equation:

(1) first we check the exactness of the differential equation.

(2) then we write the system of two differential equations that defines the function h(x,y)

h(x,y) :          = P(x,y)      ,        = Q(x,y)

(3) integrate first equation over x , here we write a unknown function g(y) instead of constant C.

h(x,y) =

(4) differentiate with respect to y , we substitute the function h(x,y) as follows;

= ( = Q(x,y)

We get,    g’(x) = Q(x,y)  - ,

Now integrate this function, we get

h(x,y) =  + g(x)

Example-1: solve (6x² - y +3)dx + (3y² -x -2)dy = 0

Sol.  First we check exactness,

(3y² -x -2) = -1        

(6x² - y +3) = -1        

Hence this is an exact differential equation.

Write the system of equation to find h(x,y),

= P(x,y) = (6x² - y +3)

= Q(x,y) = (3y² -x -2)

Integrate the first eq. w.r.t. x ,    assume y is constant,

h(x,y)  =   dx

=

= 2x³ - xy +3x +g(y)

Here we have continuous differentiable function g(y) instead of C.

Now,

=  (2x³ - xy +3x +g(y))

-x + g’(x) = 3y² - x – 2

We get,

g’(x) = 3y² – 2

Now integrate,

g(x) = = y³ - 2y

So,

h(x,y) = 2x³ -xy +3x +y³ -2y,

The general solution is defined by the following expression,

2x³ -xy +3x +y³ -2y = C

Example-2: 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.

Bernoulli’s Equation:

A Bernoulli differential equation is an equation of the form

+ p(x)y = q(x) yⁿ

When n = 1 or 0 , a Bernoulli eq. Reduces to a linear equation.

If n is real number then we make substitution,

Z =  y¹‾ⁿ

Example-3:  Solve + xy = xy²

Solution :  Here , n =2, so me make substitution,

z =y¹‾² =  y‾¹

y =

y’ = 

Now substitute these equations,

=

Or    z’ – zx = -x  -----------(1)

This is the linear form of unknown function f(z),

Integrating factor is,

I(x) = =

Multiply (1) by I(x), we get

- x -x

Or                     = -x

Integrating both sides,

  = +c

Z(x) = c

The solution for the original diff. Eq.

y = .

1.3. Application to electric circuits:

The basic equation of current I in a simple RLC, is given as below,

Where,   R = resistance  ,    L= inductor ,       E = electromotive force

+ =

For an RC circuit which consist resistance and capacitance C and no inductance , then the equation for the electric charge q on the capacitor is given by,

+ =

The relationship between q and I  is given by,

I =