Question

if t(t) is a solution of (1+t)dy/dt-ty=1 and y(0)=-1 then y(1) is equal to

Answer 1

this is a first order differential equation:\[(1+t)y'-ty=-1\]re-arrange into standard form:\[y'-\frac{t}{1+t}y=-\frac{1}{1+t}\]so the integrating factor is:\[u(t)=e^{\int -\frac{t}{1+t}dt}=e^{\ln (1+t)-t}=\frac{e^{\ln(1+t)}}{e^t}=\frac{1+t}{e^t}\]use this to re-write expression as:\[(u(t)y)'=\frac{1}{1+t}*u(t)=\frac{1}{1+t}*\frac{1+t}{e^t}=\frac{1}{e^t}\]integrate both sides:\[u(t)y=\int \frac{1}{e^t}dt=-\frac{1}{e^t}\]so:\[\frac{1+t}{e^t}y=-\frac{1}{e^t}\]\[y=-\frac{1}{1+t}\]
when t=0, y=-1 which meets the requirement
when t=1,\[y=-\frac{1}{2}\]

Answer 2

sorry - ignore the minus sign on the RHS of the first two equations I typed up.

Answer 3

they should have been:\[(1+t)y'-ty=1\]\[y'-\frac{t}{1+t}y=\frac{1}{1+t}\]