What is a general solution to the differential equation #dy/dt+3ty=sint#?

#dy/dt+3ty=sint#

we can make the LHS exact by using an integrating factor #xi(t) = e^(int 3t dt) = e^((3t^2)/2)#

#xi (dy/dt+3ty)=xi sint#

#implies e^((3t^2)/2)dy/dt+ e^((3t^2)/2)3ty= e^((3t^2)/2) sint#

this is helpful because the LHS is now:

# e^((3t^2)/2)dy/dt+ e^((3t^2)/2)3ty = d/dt( y e^((3t^2)/2) )#

So # d/dt( y e^((3t^2)/2) ) = e^((3t^2)/2) sint#

and now the RHS looks horrible

#implies y e^((3t^2)/2) =int e^((3t^2)/2) sint dt#

which cannot be solved using elementary functions

The differential equation is linear non homogeneous so the solution is composed as

#y = y_h + y_p#

where

#(dy_h)/(dt)+3t y_h = 0#
#(dy_p)/(dt)+3t y_p = sint#

The homogeneous solution can be easily obtained. So grouping variables

#dy_h/y_h=-3tdt#

giving

#y_h = C e^(-3/2t^2)#

The particular solution can be obtained adopting the so called constants variation technique due to Lagrange. Posing

#y_p = C(t)e^(-3/2t^2)# and substituting in the particular equation, giving

#d/(dt)C(t) = e^(-3/2t^2)sint#

The next step is the #C(t)# determination.

#C(t) = int e^(-3/2t^2)sint dt#. We will solve this integral solving instead

#-i "Im"[int e^(-3/2t^2)(cost+isint)dt]#

#int e^(-3/2t^2)(cost+isint)dt = int e^(-3/2t^2+i t) dt#

Choosing now #alpha, beta, gamma# such that

#(alpha t + i beta)^2=-(-3/2t^2+i t + gamma)#

#alpha = sqrt[3/2], beta = -1/sqrt[6], gamma = 1/6#

and making #y = alpha t + beta# we have

#alpha dt = dy# and the integral reads

#int e^(-3/2t^2+i t) dt equiv e^(gamma)int e^(-y^2) dy = e^gamma sqrt(pi)/2"erf"(y)#

so, finally

#C(t) = i "Im"[e^gamma sqrt(pi)/2"erf"(i/sqrt[6] - sqrt[3/2] t)]#

and

#y = i C_1"Im"[e^gamma sqrt(pi)/2"erf"(i/sqrt[6] - sqrt[3/2] t)]e^(-3/2t^2)#