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Official PYQ · 1995

GATE CS 1995 Q2.18 — Calculus: Limits, Continuity, Differentiability

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The solution of differential equation y′′+3y′+2y=0y''+3y'+2y=0 is of the form

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Answer: C

To find the general solution of the second-order linear homogeneous differential equation with constant coefficients:

y′′+3y′+2y=0y'' + 3y' + 2y = 0

We first write its characteristic (auxiliary) equation by substituting y=erxy = e^{rx}:

r2+3r+2=0r^2 + 3r + 2 = 0

Factoring the quadratic equation, we get:

(r+1)(r+2)=0(r + 1)(r + 2) = 0

This gives the roots:

r1=−1andr2=−2r_1 = -1 \quad \text{and} \quad r_2 = -2

Since the roots are real and distinct, the general solution is of the form:

y(x)=C1e−x+C2e−2xy(x) = C_1 e^{-x} + C_2 e^{-2x}

where C1C_1 and C2C_2 are arbitrary constants. This matches the expression in Option C.

Note: While Option D (C1e−2x+C2e−xC_1 e^{-2x} + C_2 e^{-x}) is mathematically identical due to the arbitrary nature of the constants, Option C is the standard representation matching the given key.

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