Solve the initial value problem y' + p(t)y = 0, y(0) = 1 where p(t) = {1, 0<= t <= 1 and 1, t > 1}
We have the function p(t), defined by
p(t) = {1, 0<= t <= 1 and 1, t > 1}
which, in effect, is simply,
p(t) = 1, t >= 0.
The ODE then is,
y^'=-y,t≥0
Integrating both sides,
∫dy/y=-∫1 dt
ln(y)-ln(k)=-t
y/k=e^(-t)
y=k.e^(-t),t≥0
Initial condition
y(0)=1=k∙1=>k=1
Answer: y=e^(-t), t≥0