GATE 2015 EE – SET 1 – Question 27

Solution :

y = $$C_{1}e^{-2t} + C_{2}e^{-3t}$$

y(0) = 2, y = 2 , x = 0 ⇒ C_{1} + C_{2} ……….(i)

$$\frac{d^{2}y}{dt^{2}} + 5 \frac{dy}{dt}$$ + 6y = 0

($$D^{2}$$+ 5D + 6)y = 0

$$D^{2}$$+ 5D + 6 = 0

D = -2 ,-3

y(1)=-$$\left ( \frac{1-3e}{e^{3}} \right ) $$

y = -$$\left ( \frac{1-3e}{e^{3}} \right ) $$ , x = 1

-$$\left ( \frac{1-3e}{e^{3}} \right )$$ =$$ C_{1}e{-2} + C_{2}e{-3}$$
– 1 + e = $$c_{1}e^{1} +c_{2}$$

Solving (1) and (2) we have ,

$$C_{1}$$ = 3 ,$$C_{2}$$ = -1

Particular solution:  y = 3$$ e^{-2t} – e^{-3t}$$
$$\frac{dy}{dt}$$ = -$$e^{-2t} + 3e^{-3t}$$
$$\frac{dy}{dt}$$(0) = -6 + 3 = -3