Question

Y⁽³⁾ + 2y" – y' – 2y = 0; y(0) = 1, y'0) = 2, y"0) = 0; y₁ = eˣ, y₂ = e-ˣ, y3 = e⁻²ˣ

Answer 1

The solution to the given differential equation is
`\(y(t) = cx_(1)e^t + cx_(2) e^(-t) + cx_(3) e^(-2t)\), where \(cx_(1) \), \(cx_(2) \), and \(cx_(3) \)` are constants determined by the initial conditions.

Step-by-step explanation:

The given differential equation
`\(y''' + 2y'' - y' - 2y = 0\)` is a third-order linear homogeneous differential equation. The general solution to such an equation is a linear combination of the complementary solutions. In this case, the complementary solutions are
`\(y_(1) = e^t\), \(y_(2) = e^(-t)\), and \(y_(3) = e^(-2t)\)` based on the characteristic equation.

The general solution is given by
`\(y(t) = c_(1) y_(1) + c_(2) y_(2) + c_(3) y_(3) \), where \(c_(1) \), \(c_(2) \\} \), and \(c_(3)\)` are constants to be determined. To find these constants, we use the initial conditions
`\(y(0) = 1\), \(y'(0) = 2\), and \(y''(0) = 0\).`Evaluating these conditions gives a system of three equations. Solving this system, we find
`\(c₁ = (1)/(2)\), \(c₂ = (3)/(2)\), and \(c₃ = -(1)/(2)\).`

Substituting these values back into the general solution, we get the particular solution
`\(y(t) = (1)/(2)e^t + (3)/(2)e^(-t) - (1)/(2)e^(-2t)\).` Therefore, the final answer is
`\(y(t) = (1)/(2)e^t + (3)/(2)e^(-t) - (1)/(2)e^(-2t)\).`