Solve the given initial-value problem. \[ y^{\prime \prime \prime}+10 y^{\prime \prime}+25 y^{\prime}=0, y(0)=0, y^{\prime}(0)=1, y^{\prime \prime}(0)=-6 \] \( y(x)=\square \)

To solve the initial-value problem given by the differential equation \[ y^{\prime \prime \prime}+10 y^{\prime \prime}+25 y^{\prime}=0, \] we start by rewriting the equation as follows: \[ y^{\prime \prime \prime} + 10y^{\prime \prime} + 25y^{\prime} = 0. \] This is a third-order linear homogeneous differential equation. We can find its characteristic equation by assuming a solution of the form \( y = e^{rx} \), which leads to: \[ r^3 + 10r^2 + 25r = 0. \] Factoring out an \( r \), we have: \[ r(r^2 + 10r + 25) = 0. \] This gives us one root \( r = 0 \). For the quadratic part \( r^2 + 10r + 25 = 0 \), we can factor it as: \[ (r + 5)^2 = 0, \] yielding a double root \( r = -5 \). Therefore, the general solution for the differential equation is: \[ y(x) = C_1 + C_2 e^{-5x} + C_3 x e^{-5x}. \] To find \( C_1 \), \( C_2 \), and \( C_3 \), we need to use the initial conditions: \( y(0) = 0 \), \( y'(0) = 1 \), and \( y''(0) = -6 \). Calculating \( y(0) \): \[ y(0) = C_1 + C_2 + 0 = 0 \implies C_1 + C_2 = 0 \implies C_2 = -C_1. \] Now, let's calculate \( y'(x) \): \[ y'(x) = -5C_2 e^{-5x} + C_3 e^{-5x} - 5C_3 x e^{-5x}. \] Evaluating \( y'(0) \): \[ y'(0) = -5C_2 + C_3 = 1. \] Substituting \( C_2 = -C_1 \): \[ 5C_1 + C_3 = 1 \implies C_3 = 1 - 5C_1. \] Next, we calculate \( y''(x) \): \[ y''(x) = 25C_2 e^{-5x} - 10C_3 e^{-5x} + 5C_3 x e^{-5x}. \] Evaluating \( y''(0) \): \[ y''(0) = 25C_2 - 10C_3 = -6. \] Substituting \( C_2 = -C_1 \) and \( C_3 = 1 - 5C_1 \): \[ 25(-C_1) - 10(1 - 5C_1) = -6, \] \[ -25C_1 - 10 + 50C_1 = -6 \implies 25C_1 - 10 = -6 \implies 25C_1 = 4 \implies C_1 = \frac{4}{25}. \] Thus, \( C_2 = -\frac{4}{25} \) and \( C_3 = 1 - 5 \cdot \frac{4}{25} = 1 - \frac{20}{25} = \frac{5}{25} = \frac{1}{5} \). Finally, substituting the values of \( C_1 \), \( C_2 \), and \( C_3 \) back into the general solution, we have: \[ y(x) = \frac{4}{25} - \frac{4}{25} e^{-5x} + \frac{1}{5} x e^{-5x}. \] Thus, the final solution to the initial-value problem is: \[ y(x) = \frac{4}{25}(1 - e^{-5x}) + \frac{1}{5}x e^{-5x}. \]