Solve for ( v(t) ) using initial condition (usually ( v_0 ) at ( t=0 )). The manual then often uses ( v = dx/dt ) to find ( x(t) ) with a second integration.
Set up the differential equation. [ \fracdvdt = 2 - 0.1v ] Solve for ( v(t) ) using initial condition
They forget the ( dv = -10, du ) substitution or try to integrate without separating variables first. The solutions manual shows this substitution explicitly. [ \fracdvdt = 2 - 0
Integrate both sides. The manual’s key move: substitute ( u = 2 - 0.1v ), so ( du = -0.1, dv ) → ( dv = -10, du ). [ \int \frac-10, duu = \int dt ] [ -10 \ln|u| = t + C ] [ -10 \ln|2 - 0.1v| = t + C ] The manual’s key move: substitute ( u = 2 - 0
That’s a classic variable acceleration problem. The solutions manual for Ch. 11 is correct, but let me clarify the logic.