Question

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

Answer 1

cannot be solved using elementary functions

Explanation:

`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

Answer 2

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

Explanation:

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)`