Let [math: y] be a twice differentiable function in [math: x] such that [math: y″+4y=4y′]y’’+4y=4y’, [math: y(0)=2] and [math: y′(0)=4]y’(0)=4. Compute [math: y(1)]. (Correct the answer to 2 decimal places.)

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We are given a twice differentiable function y(x) satisfying y'' + 4y = 4y', with initial conditions y(0) = 2 and y'(0) = 4, and we need y(1) to two decimals.

First, rewrite the differential equation in standard linear form: y'' - 4y' + 4y = 0. This has characteristic equation r^2 - 4r + 4 = 0, which factors as (r - 2)^2 = 0. Hence there is a repeated root r = 2.

Therefore the general solution is y(x) = (A + Bx) e^{2x}, where A and B are constants to be determined from initial conditions.

Apply the initial condition y(0) = 2: y(0) = (A + B·0) e^{0} = A = 2, so A = 2.

Next, compute y'(x). Differentiating y(x) = (A + Bx) e^{2x} gives y'(x) = B e^{2x} + (A + Bx)·2 e^{2x} = e^{2x} [B + 2A + 2Bx].

Use y'(0) = 4: y'(0) = e^{0} [B + 2A + 0] = B + 2A. With A = 2, this becomes B + 4 = 4, so B = 0.

Thus the particular solution is y(x) = (2 + 0·x) e^{2x} = 2 e^{2x}.

Finally, evaluate at x = 1: y(1) = 2 e^{2} ≈ 2 × 7.389056... ≈ 14.778112..., which rounds to 14.78 to two decimal places.

Hence y(1) ≈ 14.78.

_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″+4y=4y′]y’’+4y=4y’, [math: y(0)=2] and [math: y′(0)=4]y’(0)=4. Compute [math: y(1)]. (Correct the answer to 2 decimal places.)

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