Ordinary Differential Equation

An ordinary differential equation involves function and its derivatives. It contains only one independent variable and one or more of its derivatives with respect to the variable.

The order of ordinary differential equations is defined as the order of the highest derivative that occurs in the equation. The general form of n-th order ODE is given as

F(x, y, y’,…., yn ) = 0

An n-th order ordinary differential equations is linear if it can be written in the form;

a0(x)yn + a1(x)yn-1 +…..+ an(x)y = r(x)

The function aj(x), 0 ≤ j ≤ n are called the coefficients of the linear equation. The equation is said to be homogeneous if r(x) = 0. If r(x)≠0, it is said to be a non- homogeneous equation. 

Applications

1. It helps to predict the exponential growth and decay, population and species growth. 

2. Modelling the growth of diseases

3. Describes the movement of electricity

4. Describes the motion of the pendulum, waves

5. Used in Newton’s second law of motion and Law of cooling.

EXACT DIFFERENTIAL EQUATION

CONDITION: Mdx+Ndy=0 is an exact differential eqn 

                    if ӘM=ӘN

                       Әy    Әx

du =Әudx+Әudy = Mdx+Ndy       where  Әu/Әx=M &Әu/Әy=N

        Әx      Әy

        

Example 1:

 Solve

(3x²y³ − 5x⁴) dx + (y + 3x³y²) dy = 0

In this case we have:

• M(x, y) = 3x²y³ − 5x⁴
• N(x, y) = y + 3x³y²

We evaluate the partial derivatives to check for exactness.

• ∂M∂y = 9x²y²
• ∂N∂x = 9x²y²

They are the same! So our equation is exact.

Now we want to discover I(x, y)

Let's do the integration with x as an independent variable:

I(x, y) = ∫M(x, y) dx

= ∫(3x²y³ − 5x⁴) dx

= x³y³ − x⁵ + O

Note: O is our version of the constant of integration "C" because (due to the partial derivative) we had y as a fixed parameter that we know is really a variable.

So now we need to discover O

At the very start of this page we said that N(x, y) can be replaced by ∂I∂y, so:

∂I∂y = N(x, y)

Which gets us:

3x³y² + dOdy = y + 3x³y²

Cancelling terms:

dOdy = y

Integrating both sides:

f(y) = y²2 + C

We have f(y). Now just put it in place:

I(x, y) = x³y³ − x⁵ + y²2 + C

and the general solution (as mentioned before this example) is:

I(x, y) = C

So we get:

x³y³ − x⁵ + 0.5y²= C

EXAMPLE 2 :  SOLVE THE FOLLOWING EQUATION FOR EXACTNESS & FIND SOLUTION

SECOND ORDER LINEAR DIFFERENTIAL EQUTION

THESE ARE EQUATIONS GIVEN BY THE FORM : y"+py'+qy=r(x)

THE second order LDE ARE OF 2 TYPES:

• HOMOGENOUS
• HETROGENOUS 

A) HOMOGENOUS ODE 

FOR SOLVING HOMOGENOUS LDE , FIRSTLY FORM THE AUXILLARY EQUATION (AE)& SOLVE IT 

DEPENDING ON THE NATURE OF  THE ROOT OF THE AE, SOLUTION  IS GIVEN

 

CASE 1 : REAL & DISTINCT ROOTS(Q,W)

 x(t)=c1e^Qt+c2e^Wt

CASE 2 :  REAL & EQUAL ROOTS(Q,Q)

x(t)=(c1+c2t)e^Qt

CASE 3 : COMPLEX & DISTINCT ROOT (Q=a±ib)

x(t)=e^Qt[C1cosbx+c2sinbx]

EXAMPLES

Example 1 :

 Solve the IVP

y′′−10y′+29=0given y(0)=1,y′(0)=3

 

Solution : 

The characteristic equation is r2−10r+29=0

 & Auxiliary equation is D²-10D=0

which has roots

r=5+2iandr=5−2i.

 

The general solution for complex root is

y=e^5t[c1cos(2t)+c2sin(2t)].

 

We use the initial values to find the constants. Plug in y(0)=1

 

1=1[c1(1)+c2(0)]

so that c1=1

 We have

y′=5e^5t[cos(2t)+c2sin(2t)]+e^5t[−2sin(2t)+2c2cos(2t)]

Plugging in y′(0)=3

3=5[1+0]+1[0+2c2]=5+2c2.

Hence c2=−1

The final solution is y=e^5t[cos(2t)−sin(2t)].

Example 2 : 

Solve y''+y=r(x)

 Solution :  Auxiliary equation : D²+1=0

 => D=±i

Final solution: y=C1cosx + C2sinx

Example 3 :

 Solve y''+y'- 2y =0 

 Solution : Auxiliary equation

       D²+D-2=0  

     => D= 1,-2

Here ,General solution for real & distinct root :       c1e^x+c2e^-²x

TO GET A BETTER HOLD ON THE TOPIC , U CAN WATCH THE ATTACHED VIDEO LECTURE BY DR. TREFOR BAZETT