Solve the initial value problem?

Question

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

Answer 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 = e^2t

For 0≤t≤1,

y = e^-2t [∫e^2t g(t) dt + c]

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

 

 

 

Answer 2

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

e^2x y' + 2e^2x = e^2x

e^2x y = ∫ e^2x dx = e^2x/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.