00:01
The given differential equation is t squared y double dice plus t times of y dies plus t squared negative 1 over 4 of y is equal to 0.
00:15
Also if one solution y1 is given as t to the power negative 1 over 2, cos of t.
00:28
So we need to find the solution y of t.
00:33
So given differential equation can be written as y double x t over t square y -dice plus 1 negative 1 over 4 t squared of y is equal to 0 where t is written it implies y double x plus y dash over t plus 1 negative 1 over 4 t square of y is equal to 0 where t is greater than 2.
01:11
So from this differential equation, we have the function p of t is equal to 1 over 2.
01:19
So we need to find the second solution y2 is equal to y1 the integration of t to the part negative of p d t over the first solution that is y1 to whole square now substitute the value of p that is y1 integration of that is e to the part negative 1 over t t t t t t 4 over y 1 that is t to the part negative 1 over 2 of t the whole square t is equal to t to the part negative 1 over to the cause of t the part integration of e to the part integration of 1 over t that is a negative of natural log of t over t inverse cost square t 2 this is equal to 3 to the power negative 1 over 2 of t the integration of 1 over t over 1 over t over 2 1 over t off 2 so this is equal to 1 over 3 1 over 2 get cancelled out we have 3 to the power negative 1 over 2 pos of t in bracket in integration, sorry, in integration, 1 over cost, that is t squared t t t, t is equal to t to par negative 1 over 2, cost of t, multiply to the integration of take square, which is equal to 10 of t.
03:21
Now, t to 2 2 2 2 % of t, 10 is 0 % of t, 10 is sine t over 2, plus of t...