00:01
Given differential equation dy by dt is t plus 9y whole square given initial condition y of 0 is 6.
00:12
Let us solve this by separating variables and before that let us make a substitution that u equal to t plus 9y.
00:23
So if u equal to t plus 9y, let us find what is du by dt.
00:29
Du by dt is 1 plus 9 into dy by dt.
00:36
So dy by dt from this expression is it is 1 by 9 into du by dt minus 1.
00:49
Now let us convert the given equation 1 using plug -in 2 in 1.
00:56
So dy by dt is 1 by 9 du by dt minus 1 equal to u square.
01:05
Now this can be written as du by dt is 9u square plus 1.
01:14
Now separating variables it is du by 9u square plus 1 equal to dt.
01:21
Now integrating both sides we get integral of du by 9u square plus 1 is integral of dt.
01:34
So integrating we get it is we apply the formula integral of dx by a square plus x square is 1 by a tan inverse of x by a.
01:46
Let us apply this formula here.
01:47
So this will be since we have 9 outside let us take 1 by 9 common.
01:55
So it is integral of du by u square plus 1 by 9 equal to integral of dt is t.
02:06
So now 1 by 9, so 1 by 9 integral of du by u square plus 1 by 9 t plus some integral constants.
02:16
So this is going to be using the formula, using this formula.
02:20
This is going to be 1 by 9 into 3 tan inverse of 3u equal to t plus c...