Solve the differential equation y^(4) + 4y'' + 4y = 0 with initial conditions y(0) = 1, y'(0) = 0, y''(0) = -2, y'''(0) = 0.

April 10, 2026Mathematics

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April 10, 2026

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1. Solve the following differential equation: $y^{(4)} + 4y'' + 4y = 0$ Find the general solution and then determine the particular solution satisfying $y(0) = 1$ , $y'(0) = 0$ , $y''(0) = -2$ , $y'''(0) = 0$ .

Step 1: Find the characteristic equation. The given differential equation is $y^{(4)} + 4y'' + 4y = 0$ . The characteristic equation is obtained by replacing derivatives with powers of $r$ : $r^4 + 4r^2 + 4 = 0$

Step 2: Solve the characteristic equation. Let $u = r^

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1. Solve the following differential equation: y^(4) + 4y'' + 4y = 0 Find the general solution and then determine the particular solution satisfying y(0) = 1, y'(0) = 0, y''(0) = -2, y'''(0) = 0.

Asked on April 10, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

ScanSolve AI AnswerAnswered on April 10, 2026

1. Solve the following differential equation: y^(4) + 4y'' + 4y = 0 Find the general solution and then determine the particular solution satisfying y(0) = 1, y'(0) = 0, y''(0) = -2, y'''(0) = 0. Step 1: Find the characteristic equation. The given differential equation is y^(4) + 4y'' + 4y = 0. The characteristic equation is obtained by replacing derivatives with powers of r: r^4 + 4r^2 + 4 = 0 Step 2: Solve the characteristic equation. Let $u = r^