00:01
The given differential equation is y' ' plus 9y.
00:05
This will be equal to e raised to power minus twice f.
00:09
Now we have to solve this homogeneous ordinary differential equation and find the general solution.
00:16
So this is the given homogeneous ordinary differential equation.
00:20
Then general homogeneous solution is given by the characteristic equation is r square plus 9 this will be equal to 0 which has a complex roots that implies r is equal to plus minus plus minus thrice of i.
00:35
The general homogeneous solution therefore will be yh of x will be equal to c1 into cos of thrice x plus c2 into sin of 3x.
00:48
So this is our general homogeneous solution.
00:53
Then particular solution.
00:54
We will use the method of undetermined coefficients.
00:58
Let guess a particular solution of the form yp of x.
01:02
This will be equal to a into e raised to power minus thrice f that is this rhs for given.
01:09
So this will be substitute this into the differential equation.
01:13
We get that 4 into a plus 9 into a this will be equal to 1.
01:22
So this will gives us solving for a we get 13 a will be equal to 1.
01:28
A will be equal to 1 divided by 13.
01:32
So the particular solution is given by here yp.
01:38
So yp of x it is equal to 1 divided by 13 into e raised to power minus thrice x.
01:46
Then general solution.
01:49
Then general solution is given by the sum of homogeneous and the particular solution that means this sum of this two.
01:58
So this y of x will be equal to yh of x that is homogeneous solution plus yp of x that is particular solution.
02:10
So this will be equal to y of x is c1 cos of thrice x plus c2 into sin of thrice x plus 1 by 13 multiplied by e raised to power minus thrice x.
02:28
Now this was the path first and this is the required solution for path first of a question.
02:37
This was first part of a question and this part is very easy part since we are not provided any initial conditions.
02:46
So we cannot calculate here c1 and c2 constants but in second and third part we are provided the equation along with the initial conditions.
02:55
So for that we will apply the initial conditions and find the constants also.
03:00
Then the second and third problem will be lengthy as compared to first.
03:05
So in part number 2 we are provided the initial value problems given differential equation is 2y double dash plus 5 into y dash plus 3 into y this will equal to 0 and initial conditions as i told before is given as y of 0 is equal to 3 and y dash of 0 it is equal to minus 4.
03:28
So we will solve this homogeneous equation twice r square plus 5r plus 3 it is equal to 0 and this will implies the values of r as minus 1 comma minus 3 by 2.
03:44
Therefore the general homogeneous solution y of x is given as e raised to power minus x multiplied by c1 constant plus c2 constant e raised to power minus 3 by 2.
03:56
So this is the general homogeneous ordinary differential equations general homogeneous solution.
04:03
Now we will find the c1 c2.
04:05
Initial condition was y of 0 is equal to 3.
04:10
So we will substitute here y of 0 that is x is 0 then we get here 3 is equal to c1 plus c2 then y dash of 0 is equal to minus 4 we will substitute then we get y dash of 0 it is equal to minus c1 minus 3 by 2 into c2.
04:32
This will be equal to minus 4 since y dash of 0 is minus 4 therefore this will be the since here we can find its derivative it will be minus of c1 into this and minus of 3 by 2 into this c2...