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.?

1 Answer
Jul 3, 2018

Answer:

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

Explanation:

Question is:

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

  • For #t in [0,1]#

#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))#

  • For #t gt 1 #

#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):}#