00:01
So, the equation that has been given to us is cauchy's non -homogeneous equation.
00:14
So we will just write it once again.
00:16
It is x square d d square minus 2y is equal to 5x cube plus log of x square.
00:30
So what we will do is we will put x equals to e raised to t.
00:37
So taking log on both sides what i will get is t is equal to log of x.
00:43
So that what we get now is x multiplied by d is equal to the cursive d.
00:55
So x square capital d square is equal to cursive d in bracket cursive d minus 1.
01:03
So what is the difference between this capital d is it is d over dx and the value of cursive d it is equal to d over dt.
01:13
I am just using this cursive and simple d to signify the notions.
01:20
So then our initial equation becomes d multiplied by in bracket cursive d minus 1 minus 2 and here it is y.
01:33
So it is equal to 5 multiplied by e raised to 3t plus t square.
01:39
So it becomes d square minus d minus 2 times y is equal to 5 multiplied by e raised to 3t plus t square.
01:52
So this is our second equation.
01:55
So what we will do is the auxiliary equation out of this is d square minus d minus 2 is equal to 0.
02:04
So we will just solve this equation.
02:06
What i will do is i will just split it up.
02:08
So it becomes d square minus 2 d plus d minus 2 is equal to 0.
02:18
So i have just written minus d at minus 2 d plus d.
02:22
So what we get is minus 2 d minus 2 multiplied by d plus 1.
02:29
So is equal to 0.
02:31
So we get the value of roots as d is equal to 2 comma minus 1.
02:34
So now what are value of yh becomes c1 e raised to 2t plus c2 into e raised to minus t.
02:49
So the value of yh turns out to be c1 x square plus c2 multiplied by x raised to minus 1.
02:58
So this is our value of yh.
03:02
Similarly for the b part of the equation what we will do is the particular solution method of undetermined coefficient...