00:02
Hello everyone.
00:03
Today i will be discussing second order differential equation.
00:24
Okay, the type of equation i'm going to discuss is ay double dash plus b y -dash plus cy equals to rx.
00:32
This is a second order linear ordinary differential equation of the non -homogeneous type with constant coefficients.
00:39
I have rx which makes it non -homogeneous and a, b, c are the constants.
00:45
R is any function of x.
00:46
To solve this type of equation there are various methods.
00:51
I will be illustrating one of them through an example.
00:54
Let's consider an example.
00:57
Y double dash minus 9y equals to atex.
01:03
Let's have this simple example and i will solve this using the method of variation of parameters.
01:25
Okay, so using the method of variation of parameters.
01:29
First of all, we will find out the complementary function.
01:33
Complementary function is the solution of the corresponding homogeneous equation.
01:38
What is a homogeneous equation where rx is equal to 0? so we will simply substitute rx equal to 0.
01:47
This is the corresponding homogeneous equation.
01:50
Using this, we will find out our auxiliary equation.
01:55
Auxiliary equation is r square minus 9 r square for y double dash and minus 9 equal to 0 this gives us our roots as now we will find out its roots are equal to plus minus 3 so these are real and distinct roots so our complementary function becomes c1 e minus 3x plus c2 e 3x this is what our complementary function is which is the solution of the homogeneous equation now we will try to find out the particular solution okay to find this we will use the method of variation of parameters for variation of parameters what i will be using is the two linearly independent functions which were obtained in my complementary functions.
03:12
First one was y1 equals to e to the par minus 3x and the second one was y2 equals to e to the bar 3x.
03:21
So these are the two functions which i have already obtained in my complementary function...