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

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

Subscribe to Thesis123 YouTube Channel by Clicking ( HERE ) For  FREE Video Courses