Let [math: y] be a twice differentiable function in [math: x] such that [math: y″+5y′+6y=0]y’’+5y’+6y=0, [math: y(0)=1] and [math: y′(0)=2]y’(0)=2. Compute [math: y(1)]. (Correct the answer to 2 decimal places.)

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We are given a second-order linear ODE: y'' + 5y' + 6y = 0 with y(0) = 1 and y'(0) = 2.

First, solve the homogeneous equation by finding the characteristic roots: r^2 + 5r + 6 = 0 factors as (r + 2)(r + 3) = 0, giving r = -2 and r = -3.

This implies the general solution is a linear combination of exponentials: y(x) = A e^{-2x} + B e^{-3x}.

Next, apply the initial condition y(0) = 1. Substituting x = 0 gives y(0) = A + B = 1, so A = 1 - B.

Then use the second initial condition for the derivative. Compute y'(x) = -2A e^{-2x} - 3B e^{-3x}. At x = 0, y'(0) = -2A - 3B = 2.

Substitute A = 1 - B into the derivative condition: -2(1 - B) - 3B = 2. This simplifies to -2 + 2B - 3B = 2, i.e. -2 - B = 2, so B = -4.

Now find A = 1 - B = 1 - (-4) = 5.

With A and B known, the specific solution is y(x) = 5 e^{-2x} - 4 e^{-3x}.

Finally, evaluate at x = 1: y(1) = 5 e^{-2} - 4 e^{-3}. Numerically, e^{-2} ≈ 0.135335, so 5 e^{-2} ≈ 0.676675; e^{-3} ≈ 0.0497871, so -4 e^{-3} ≈ -0.199148; summing gives y(1) ≈ 0.676675 - 0.199148 ≈ 0.477527, which rounds to 0.48 to two decimal places.

_MATH1013_1ABCD_2025 Subsection 6.2 (closed on 22 Nov)

Let [math: y] be a twice differentiable function in [math: x] such that [math: y″+5y′+6y=0]y’’+5y’+6y=0, [math: y(0)=1] and [math: y′(0)=2]y’(0)=2. Compute [math: y(1)]. (Correct the answer to 2 decimal places.)

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