Gate 2016 EE (Set 1) Q4

4. A function y(t) , such that  y(0)=1 and y(1)=3e^(-1) , is a solution of the differential equation  (d^2 y)/(dt^2 )+2dy/dt+y=0. Then y(2) is :--------

Solution:

 (d^2 y)/(dt^2 )+2dy/dt+y=0

 D^2 y+2Dy+y=0
 (D^2+2D+1)y=0
 (D+1)(D+1)y=0
 D= -1,-1

Roots are real and repetitive.

General solution:  y=e^ax (C_1+C_2 x),where a= -1

 y(0)=1=>y=1,x=0 in the above equation gives us  C_1=1

 y(1)=3e^(-1)=>y=3e^(-1),x=1 in the above equation,gives us  C_2=2

Particular solution: y=e^(-x) (1+2x)

 y(2)=e^(-2) (1+4)=5e^(-2)

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