GATE 2024 IN Q.64: solution and answer

GATE 2024 · INQ.64 · 2 marks · NAT

The solution of an ordinary differential equation $y^{'''} + 3 y^{''} + 3 y^{'} + y = 30 e^{- t}$ is
$y (t) = (c_{0} + c_{1} t - c_{2} t^{2} + c_{3} t^{3}) e^{- t}$
Given $y (0) = 3$ , $y^{'} (0) = - 3$ and $y^{''} (0) = - 47$ , the value of $(c_{0} + c_{1} + c_{2} + c_{3})$ is ________ (rounded off to nearest integer).

Note: $y^{'''} = d^{3} y / d t^{3}$ , $y^{''} = d^{2} y / d t^{2}$ , $y^{'} = d y / d t$ and $c_{0}, c_{1}, c_{2}, c_{3}$ are constants.

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Answer and solution

ANSWER

$c_{0} + c_{1} + c_{2} + c_{3} = 3 + 0 + 25 + 5 = 33$

Official answer: 33Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

Characteristic equation: $r^{3} + 3 r^{2} + 3 r + 1 = (r + 1)^{3} = 0$ , so $r = - 1$ is a triple root. The complementary function is $(A + B t + C t^{2}) e^{- t}$ . This gives the $c_{0}$ , $c_{1}$ and $c_{2}$ terms of the given form, with $C = - c_{2}$ .
Particular integral: the forcing term $e^{- t}$ corresponds to a root of multiplicity 3, so try $y_{p} = K t^{3} e^{- t}$ . The exponential shift rule $(D + 1)^{3} [e^{- t} f (t)] = e^{- t} D^{3} f (t)$ gives $(D + 1)^{3} [K t^{3} e^{- t}] = K e^{- t} D^{3} (t^{3}) = 6 K e^{- t} = 30 e^{- t} \Rightarrow K = 5$ So the ODE alone fixes $c_{3} = 5$ . The initial conditions cannot change it, because $D^{3}$ of $c_{0} + c_{1} t - c_{2} t^{2}$ is zero.
Write $y = u e^{- t}$ with $u = c_{0} + c_{1} t - c_{2} t^{2} + c_{3} t^{3}$ . Then $y^{'} = (u^{'} - u) e^{- t}$ and $y^{''} = (u^{''} - 2 u^{'} + u) e^{- t}$ . At $t = 0$ : $u (0) = c_{0}$ , $u^{'} (0) = c_{1}$ , $u^{''} (0) = - 2 c_{2}$ .
$y (0) = c_{0} = 3$ .
$y^{'} (0) = c_{1} - c_{0} = - 3 \Rightarrow c_{1} = 0$ .
$y^{''} (0) = - 2 c_{2} - 2 c_{1} + c_{0} = - 2 c_{2} - 0 + 3 = - 47 \Rightarrow c_{2} = 25$ . Note that the question prints a minus sign in front of $c_{2} t^{2}$ .
Check: $u = 3 - 25 t^{2} + 5 t^{3}$ gives $u^{'''} = 30$ , so the left side is $e^{- t} \cdot 30 = 30 e^{- t}$ . Also $y (0) = 3$ , $y^{'} (0) = 0 - 3 = - 3$ and $y^{''} (0) = - 50 - 0 + 3 = - 47$ .
Sum: $c_{0} + c_{1} + c_{2} + c_{3} = 3 + 0 + 25 + 5 = 33$ .

KEY IDEA

The operator is $(D + 1)^{3}$ , so substituting $y = u e^{- t}$ turns the ODE into $u^{'''} = 30$ . That fixes $c_{3} = 5$ straight away. The three initial conditions, taken through $y^{'} = (u^{'} - u) e^{- t}$ and $y^{''} = (u^{''} - 2 u^{'} + u) e^{- t}$ at $t = 0$ , then give $c_{0}$ , $c_{1}$ and $c_{2}$ one after another.

WATCH OUT FOR

Missing the printed minus sign in front of $c_{2} t^{2}$ . With it, $u^{''} (0) = - 2 c_{2}$ and $c_{2} = 25$ . If you drop the sign you get $c_{2} = - 25$ and a sum of $- 17$ . Another error is to try to find $c_{3}$ from the initial conditions: they only fix $c_{0}$ , $c_{1}$ and $c_{2}$ , while $c_{3}$ comes from the forcing term.

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GATE 2024 IN Q.64: solution and answer

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