Step 1: Let the trains meet after time \( t \), having covered distances \( v_1 t \) and \( v_2 t \) respectively, where \( v_1 t \) is what train 2 still needs to cover after the meeting, and \( v_2 t \) is what train 1 still needs to cover.
Step 2: After meeting, train 1 covers the remaining distance \( v_2 t \) in 4 hours: \( v_2 t = 4 v_1 \). Train 2 covers the remaining distance \( v_1 t \) in 9 hours: \( v_1 t = 9 v_2 \).
Step 3: Multiplying these two equations together eliminates the speeds symmetrically: \( (v_2 t)(v_1 t) = (4v_1)(9v_2) \), giving \( v_1 v_2\, t^2 = 36\, v_1 v_2 \), so \( t^2 = 36 \) and \( t = 6 \).
Step 4: Substitute back into \( v_2 t = 4v_1 \): \( 6 v_2 = 4 v_1 \), so \( \frac{v_1}{v_2} = \frac{6}{4} = \frac{3}{2} \).
\[ \boxed{3:2} \]