How to solve Second order Differential Equations by tutorcircleteam

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									   How to solve Second order Differential Equations
How to solve Second order Differential Equations

A second order differential equation is of the form A(x) d2x/dy2 + B(x) dy/dx +
R(x)y = G(x) ….........equation(1) Where A , B , R , and G are continuous
functions.

This type of equations is used in study of motion of spring. When value of G(x)
= 0 , for all x , in equation 1 then this type of equation is called 'Homogenious
equations'. A(x) d2x/dy2 + B(x) dy/dx + R(x)y = 0 , is a homogeneous
equation.

Example of a homogeneous linear equation – Let us take a Homogeneous
Linear equation y” + y' – 6y = 0. Solution for the above equation is shown by the
steps below:- Case 1: When y1(x) and y2(x) are both solutions of the linear
homogeneous equation and c1 and c2 are any constants.
                     Know More About Rational Numbers Reciprocal Worksheet
Step 1: Find Auxiliary Equation which is r2 + r – 6 = 0? => (r – 2)(r + 3)=0 so
roots of the equation are r = 2 and r = -3.

Step 2: Find General Solution y = c1e2x + c2e-3x It is calculated by using
formula y = c1y1(x) + c2y2 (x) Case 2 : When roots are real and distinct.

Example: 3d2y / dx2 + dy / dx – y = 0? Solution)

Step 1: Find the auxiliary equation that is 3r2 + r – 1 = 0 => r = (1 + √13 ) / 6 , r
= (1 - √13) / 6

Step 2 : Find the general solution. Which is y = c1e(-1 + √13)x / 6 + c2e(-1 -
√13)x / 6.

Case 3: When roots are real and equal

Example: 4y” + 12y' + 9y = 0

Step 1: Find auxiliary equation that is 4r2 + 12r + 9 = 0.       => ( 2r + 3 )2 = 0,
so root r = -3/2



                       Learn More About Rational Number Reciprocal Worksheets
Step 2: Find the general solution. y = c1e-3x/2 + c2xe-3x/2 this is calculated by
formula c1erx + c2xerx. Case 4: When roots r1 and r2 are complex numbers
are complex numbers.

Example: y” - 6y' + 13y = 0

Step 1: Find the Auxiliary Equation that is r2 + 6r + 13 = 0. Therefore roots r1 = [
6 ± ( √ 36 – 52 ) ] / 2 = 3 ± 2i

Step 2 : Find the general solution which is, y = e3x ( c1 cos 2x + c2 sin 2x )
this is calculated by formula y = eαx ( c1 cos βx + c2 sin βx ).
   Thank You




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