Differential Linear Equations?

Question

Find the general solution of this differential equation. t*(dy/dt) + 2*y = sin(t) for t > 0. Thanks.

Answer 1

Divide by t and get
dy/dt + 2 · y/t = sin(t)/t

You can solve such an ordinary DE of the form
dy/dt + p(t) · y = q(t)
with the integrating factor:
µ = exp( ∫ p(t) dt )
the general solution is
y =( ∫ µ · q(t) dt + C ) / µ

For this problem
µ = exp( ∫ 2/t dt ) = exp ( 2·ln(t)) = t²

y = ( ∫ t² · sin(t) / t dt + C ) / t²
= ( ∫ t · sin(t) dt + C ) / t²
= (sin(t) - t · cos(t) + C ) / t²
= sin(t)/t² - cos(t)/t + C/ t²

Answer 2

Only if you mow my lawn.