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Solve the initial value problem?

Solve the initial value problem y'+2y=g(t), y(0)=0, where g(t)={1, 0≤t≤1, 0, t>1.

This is again a typical first-order linear differential equation: y' + Py = Q

y = (1/e∫P)(∫e∫P Q dx + c) <==Formula of the solution for any first-order linear differential equation.

P = 2

e∫P = e2t

For 0≤t≤1,

y = e-2t [∫e2t g(t) dt + c]

y = e-2t [(1/2)e2t + c]

y = (1/2) + ce-2t

Plug in y(0) = 0,

0 = (1/2) + c

y = (1/2)(1- e-2t) <==Answer 1

For t>1,

y = ce-2t

Since y(1) = (1/2)(1- e-2), (1/2)(1- e-2) = ce-2

c = (1/2)(e2 - 1)

y = (1/2)(e2 - 1)e-2t <==Answer 2

On interval [0,1] we have y' + 2y = 1. Multiplying by integrating factor μ = e2x gives

e2x y' + 2e2x = e2x

e2x y = ∫ e2x dx = e2x/2 + C

y = 1/2 + Ce-2x

Using the initial value, you have y(0) = 1/2 + C = 0 so that C = -1/2.

Thus you have y = (1 - e-2x)/2.

On the interval (1,∞) you have y' = -2y which is easily solved as y = Ce-2x.

If we want a continuous function at x = 1, we have (1 - e-2)/2 = Ce-2.

This makes C = (e2 - 1)/2 so that y = (e2 - 1)e-2x/2.

Note, though the function we get is continuous at x = 1, it is not differentiable at that point.