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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

 

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