Find a continuous solution of the IVP dy/dt+y=g(t),y(0)=0 where g(t)=[2,0≤t≤1 0,t>1.?

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Answer 1 · 2018-07-03T11:49:57

$y={(2 ( 1 - e^(-t)),0≤t≤1),( 2 (e - 1)e^(-t),t>1):}$

Question is:

$dy/dt+y=g(t),y(0)=0 qquad g(t)={(2,0≤t≤1),( 0,t>1):}$

$y' + y = 2$

Applying an integrating factor: $exp(int dt)$

$e^t(y' + y) bb(= (y e^t)^' )= 2 e^t$

$y e^t = 2 e^t + C$

$y = 2 + C e^(-t)$

$y(0) = 0 = 2 + C$

$:. y_1 = 2 ( 1 - e^(-t))$

$y' + y = 0$

Separating for integration:

$dy/y = - dt qquad => ln y = -t + D$

$:. y_2 = De^(-t)$

For continuity:

$y_1(1) = y_2(1) = 2$

$implies D = 2 (e - 1)$

$:. y={(2 ( 1 - e^(-t)),0≤t≤1),( 2 (e - 1)e^(-t),t>1):}$