00:01
In this question, we have been given a differential equation t square y double dash minus t square plus 2t times y dash plus t plus 2 times of y equals to t cube, correct? and we have been given that y1 equals to t and y2 equals to t e raised to t, they are solution corresponding to the homogeneous part.
00:27
We have to find the solution for the non -homogeneous part here.
00:31
So, how to do this? so, first of all, with the help of this, i can write down my complementary function cf, it comes out to be c1 t plus c2 t e raised to t, correct? now, what i will do? now, let w be the wrong skin of solution y1 and y2, then in that case w of y1, y2, we know that it is given by y1, y2, y1 dash, y2 dash, which is equals to t and we have t e raised to t1, this is e raised to t times t plus 1, solving this we get t square e raised to t and let yp of t, which is equals to y1 t times ut plus y2 t times vt be the particular solution.
01:49
So, from this what i am going to get? we know that from the method of variation of parameter, we have u equals to integration minus yt y2, correct? r over w that is the wrong skin dt and we have the formula for v as y1 r over w dt, where we have taken r as t cube over t square, which is simply t.
02:27
So, we are getting u is equals to integral minus t raised to t and this is t over wrong skin comes out to be e square e raised to t and here we have dt.
02:38
So, this comes out to be just minus 1 dt that is equals to minus t...