"Ans. 1The given differential equation,d (y) ? 6y ? 2t ? 3 dt The above equation is of first order differential equation,Its standard form is as,' y (x)?? p(x)y q(x)The general solution of the above equation is,p() x dx ? e q() x dx ?C ? yx () ? p() x dx ? e In the given equation compare it to the standard form,p(x) ? 6, q(x) ? 2t ? 3The integrating factor,p() x t Ie ? 6t Ie ? Now,d 6y ? (y) ? 2t ? 3dt 6t Multiply e in the above equation on both side,d 6t 6t 6t 6t e .6y ?e (y) ?e .2t ?e .3 dt By the integration by parts,d 6t 6t 6t (e y)?? e .2t e .3dtNow, Solve the L.H.S.d 6t 6t 6t (e y)?? e .2t e .3dt dt 6t 6t 6t e y?? e .2t e 3dt ? 1 1 1 6t 6t 6t 6t e y ? 2( e t ? e ) ? e ?c 1 6 36 2 6t Divide both side by e ,6t 9c?? e (3t 4) 1 y ?6t 9e Now apply the initial value condition5 t ? 0 c ? At , y(0) ? 1, 1 9 1 ?66tt y ? e (e (3t ? 4) ? 5)9 Ans. 2The given equation,d ?t (y)?? 2y e cos(t) dt The I. F. for the equation,?2t Ie ?Multiply the I to the both side of the equation, and apply by parts, "