Question

The particular solution of the initial value problem given below is:

\(\frac{d^2y}{dx^2}+12\frac{dy}{dx}+36y=0\), with y(0) = 3 and \(\frac{dy}{dx}|_{x=0}=-36\)

1. (3 – 18x) e^−6x
2. (3 + 25x) e^−6x
3. (3 + 20x) e^−6x
4. (3 − 12x) e^−6x

Answer 1

Correct Answer - Option 1 : (3 – 18x) e^−6x

\(\frac{d^2y}{dx^2}+12\frac{dy}{dx}+36y=0\)

Let \(\frac{d}{dx}=D\)

The auxilliary equation can be written as:

D² + 12D + 36 = 0

(D + 6)² = 0

D = -6 and -6

Complementary solution y = (C₁ + C₂x) e^-6x

Given y(0) = 3 = (C₁ + 0) e^-0

c₁ = 3

\(\frac{dy}{dx}=(C_1+C_2x)(-6e^{-6x})+e^{-6x}C_2\)

\(\frac{dy}{dx}|_{x=0}=-36=C_1(-6)+C_2\)

C₂ = -36 + 18 = -18

∴ y = (C₁ + C₂x) e^-6x

y = (3 – 18x) e−6x