Let [math: y] be a twice differentiable function in [math: x] such that [math: y″=2y′−2y]y’’=2y’-2y, [math: y(0)=1] and [math: y′(0)=3]y’(0)=3. Compute [math: y(1)]. (Correct the answer to 2 decimal places.)

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We are given a second-order linear ODE: y'' = 2y' - 2y with initial conditions y(0) = 1 and y'(0) = 3.

First, rewrite the differential equation in standard form: y'' - 2y' + 2y = 0. The characteristic equation is r^2 - 2r + 2 = 0. Solving yields r = 1 ± i.

Thus the general solution is y(x) = e^{x}[C1 cos x + C2 sin x].

Next, apply the initial condition y(0) = 1. Since cos 0 = 1 and sin 0 = 0, we get y(0) = e^{0}[C1(1) + C2(0)] = C1, so C1 = 1.

Then compute y'(x). Differentiating y(x) = e^{x}[C1 cos x + C2 sin x] gives y'(x) = e^{x}[C1 cos x + C2 sin x] + e^{x}[-C1 sin x + C2 cos x] = e^{x}[(C1 cos x + C2 sin x) + (-C1 sin x + C2 cos x)]. Evaluate at x = 0: y'(0) = [C1(1) + C2(0)] + [-C1(0) + C2(1)] = C1 + C2. The initial condition y'(0) = 3 implies C1 + C2 = 3. With C1 = 1, we get C2 = 2.

Now compute y(1): y(1) = e^{1}[C1 cos 1 + C2 sin 1] = e[1·cos(1) + 2·sin(1)].

Using numerical values: cos(1) ≈ 0.5403023059 and sin(1) ≈ 0.8414709848, so the bracket becomes 0.5403023059 + 2×0.8414709848 ≈ 0.5403023059 + 1.6829419696 ≈ 2.2232442755. Multiplying by e ≈ 2.718281828 gives y(1) ≈ 6.042859.

Rounding to two decimal places yields 6.04.

_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″=2y′−2y]y’’=2y’-2y, [math: y(0)=1] and [math: y′(0)=3]y’(0)=3. Compute [math: y(1)]. (Correct the answer to 2 decimal places.)

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