Please show all work. Im pretty stuck on this one. Linear differential equations

Question

Please show all work. Im pretty stuck on this one.

Linear differential equations sometimes occur in which one or both of the functions p(t) and g(t) for y' + p(t) y = g(t) have jump discontinuities. If t0 is such a point of discontinuity, then it is necessary to solve the equation separately for t t0. Afterward, the two solutions are matched so that y is continuous at t0. This is accomplished by a proper choice of the arbitrary constants. The following problem illustrates this situation. Note that it is impossible also to make y' continuous at t0. Solve the initial value problem y' + p(t) y = 0, y(0) = 1 where

Explanation / Answer

if 0<=t<=1

equation is

y'+2y = 0

=>

dy/y = -2dt

=>

y = c*e^(-2t)

we have y(0) =1

=>

c = 1

=>

y = e^(-2t) for 0<=t<=1

if t>1

y'+y = 0

=>

dy/y = -t

=>

y = ce^(-t)

since y is continuous at t = 1

we have ce^(-1) = e^(-2)

=>

c = e^(-1)

=>
y = e^(-1-t) for t>1

so

y =

e^(-2t) for 0<=t<=1.

e^(-1-t) for t>1

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