00:01
Hello students, this is given to us.
00:04
We have already given one solution y1 t equals to t square.
00:16
So, the second solution let the second solution is of the form y2 t equals to v of t into y1 t.
00:33
So, therefore y2 t equals to t square into v of t.
00:39
If we differentiate it we have y2 dashed equals to t square v dashed plus 2 t v and y2 double dashed is equals to t square v double dashed plus 4 t v dashed plus 2 v.
01:03
If we substitute this into the given differential equation we have t cube into t v double dashed plus 2 v dashed equals to 0.
01:16
Let us assume a power series solution, a power series solution of the form v of t is equals to summation over n equals to 0 to infinity a n t to the power n.
01:43
So, v dash t v dash t is equals to summation over n equals to 0 to infinity n into a n into t to the power n minus 1 and v double dash t equals to summation over n equals to 0 to infinity n into n minus 1 into a n into t to the power n minus 2.
02:10
If we substitute this into this differential equation we have t to the power 4 into summation over n equals to 2 to infinity n into n minus 1 a n t to the power n minus 2 plus 2 t square to summation over n equals to 0 to infinity a n t to the power n equals to 0.
02:37
Combining this we have summation n equals to 2 to infinity n into n minus 1 a n plus 2 a n minus 2 whole to the power t to the power n equals to 0.
02:55
Since this sum is 0, since n of n into n minus 1 into a n plus 2 n minus 2 equals to 0.
03:32
So, solving this we have a n a 2 k equals to minus 1 whole to the power k 2 to the power 2 k minus 1 by 2 k whole factorial into a 0.
03:49
So, therefore we can say that v of t v of t is equals to summation over k equals to 0 to infinity a 0 into a 2 k into t to the power 2 k.
04:12
This is the second solution...