GATE 2015 EE – Question 37
A solution of the ordinary differential equation $\frac{d^2y}{dt^2} + 5\frac{dy}{dt} + 6y = 0$ is such that $y(0) = 2$ and $y(1) = -\frac{1 - 3e}{e^3}$. The value of $\frac{dy}{dt}(0)$ is ________.
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: -3
Explanation
The roots of $r^2 + 5r + 6 = 0$ are $-2$ and $-3$, so $y = Ae^{-2t} + Be^{-3t}$. From $y(0) = 2$, $A + B = 2$. The value $y(1) = \frac{3e - 1}{e^3} = 3e^{-2} - e^{-3}$ gives $A = 3$ and $B = -1$. Then $y'(0) = -2A - 3B = -6 + 3 = -3$.