# 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:$$y=e^(-x) (1+2x)$$

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