y^(''')+6y^('')+11y^(')+6y=0

Characteristic Equation: The given differential equation is a linear homogeneous ordinary differential equation with constant coefficients. The characteristic equation is found by replacing y''' with r^3, y'' with r^2, y' with r, and y with 1. Characteristic equation: r^3 + 6r^2 + 11r + 6 = 0Find Roots: We need to find the roots of the characteristic equation. We can try to factor it, looking for integer roots among the factors of 6 (the constant term). By trying possible roots, we find that r = -1 is a root because (-1)^3 + 6(-1)^2 + 11(-1) + 6 = -1 + 6 - 11 + 6 = 0.Factorization: Now we can factor out (r + 1) from the characteristic equation using polynomial division or synthetic division. (r + 1)(r^2 + 5r + 6) = 0Further Factorization: The quadratic factor can be factored further: (r + 1)(r + 2)(r + 3) = 0General Solution: The roots of the characteristic equation are r = -1, r = -2, and r = -3.The general solution to the differential equation is a linear combination of exponential functions based on the roots of the characteristic equation: y(t) = C1*e^(-t) + C2*e^(-2t) + C3*e^(-3t) where C1, C2, and C3 are constants determined by initial conditions.

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y^{\prime \prime \prime}+6 y^{\prime \prime}+11 y^{\prime}+6 y=0

Characteristic Equation: The given differential equation is a linear homogeneous ordinary differential equation with constant coefficients. The characteristic equation is found by replacing $y$ with $r^3$ , $y''$ with $r^2$ , $y'$ with $r$ , and $y$ with $1$ . $\newline$ Characteristic equation: $r^3 + 6r^2 + 11r + 6 = 0$
Find Roots: We need to find the roots of the characteristic equation. We can try to factor it, looking for integer roots among the factors of $6$ (the constant term). $\newline$ By trying possible roots, we find that $r = -1$ is a root because $(-1)^3 + 6(-1)^2 + 11(-1) + 6 = -1 + 6 - 11 + 6 = 0$ .
Factorization: Now we can factor out $(r + 1)$ from the characteristic equation using polynomial division or synthetic division. $(r + 1)(r^2 + 5r + 6) = 0$
Further Factorization: The quadratic factor can be factored further: r + \(1)(r + $2$ )(r + $3$ ) = $0$ \
General Solution: The roots of the characteristic equation are $r = -1$ , $r = -2$ , and $r = -3$ .
General Solution: The roots of the characteristic equation are $r = -1$ , $r = -2$ , and $r = -3$ .The general solution to the differential equation is a linear combination of exponential functions based on the roots of the characteristic equation: $\newline$ $y(t) = C_1e^{(-t)} + C_2e^{(-2t)} + C_3e^{(-3t)}$ $\newline$ where $C_1$ , $C_2$ , and $C_3$ are constants determined by initial conditions.