00:01
Hi, there is a question we say that find the solution of the given initial value problem ty dash plus 3y equal to t square minus t plus 5 and condition is y when x equal to 1 is 7 and t is positive.
00:22
So if we write it dividing both sides by t it will become y dash equal to 3 by t y equal to t square minus t plus 5 divided by t.
00:38
So let us write y dash as dy by dt plus 3 by t y equal to t square minus t plus 5 divided by now this is just like our linear differential equation dy by dx plus p as a function of x into y equal to q as a function of x.
01:07
So according to this our differential equation is dy by dt plus p as a function of t y equal to q as a function of t.
01:20
It is integrating factor is e raised to the power p t dt.
01:26
So e raised to the power 3 by t dt integration that would be e raised to the power 3 ln t.
01:34
So e raised to the power ln t q so t q.
01:40
Now solution to these type of differential equation y into if equal to q t into if into dt plus c where c is the constant of integration.
01:54
So y is simply y and integrating factor is t q equal to q t is this t square minus t plus 5 by t t square minus t plus 5 by t multiplied by integrating factor t q dt plus c this is t square.
02:24
So y t cube equal to t raised to the power 4 minus t cube plus 5 t square dt plus c.
02:39
Now integrating it up t raised to power 5 by 5 minus t raised to power 4 by 4 plus 5 t cube by 3 plus c y t cube...