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GATE CE 2020 Official Paper: Shift 1

Option 3 : -10e^{2t} + 10e^{3t}

CT 1: Ratio and Proportion

2536

10 Questions
16 Marks
30 Mins

__Explanation:__

\(\frac{{{d^2}x}}{{d{t^2}}} - 5\frac{{dx}}{{dt}} + 6x = 0\)

Above given equation is a **linear** differential equation of **order = 2**. Such equations are solved using CF + PI method.

Let \(D = \frac{{dx}}{{dt}}\)

\( \Rightarrow \frac{{{d^2}x}}{{d{t^2}}} - 5\frac{{dx}}{{dt}} + 6x = 0\)

⇒ D^{2}x – 5Dx + 6x = 0

⇒ (D^{2} – 5D + 6)x = 0

Thus auxiliary equation (obtained by replacing D with m) is m^{2} – 5D + 6 = 0.

Roots of above obtained auxiliary equation are-

m^{2} – 5D + 6 = 0

⇒ (m – 2) (m - 3) = 0

⇒ m_{1} = 2 or m_{2} = 3

When both roots of auxiliary equation are real and distinct

\(C.F = {C_1}{e^{{m_1}t}} + {C_2}{e^{{m_2}t}}\)

Thus general solution of above D.E is

\(x = {C_1}{e^{2t}} + {C_2}{e^{3t}}\)

Applying boundary conditions, x(0) = 0

⇒ O = C_{1} + C_{2}

⇒ C_{1} = -C_{2} ---(i)

Using initial condition \(\frac{{dx}}{{dt}}\left( 0 \right) = 10\)

\(\frac{{dx}}{{dt}} = 2G{e^{2t}} + 3{C_2}{e^{3t}}\)

⇒ 10 = 2C_{1} + 3C_{2} ---(ii)

Solving equation (i) & (ii) simultaneously,

C_{1} = -10 and C_{2} = 10

Thus, **x = -10e ^{2t} + 10e^{3t}**