Exercise 0.2.9 : Verify that x = C 1 e − t + C 2 e 2 t is a solution to x ′′ − x ′ − 2 x = 0 . Find C 1 and C 2 to solve for the initial conditions x (0) = 10 and x ′ (0) = 0 .

Answer:Since 2 and -1 are eigenvalues of the differential equation,[tex]x(t) = c_{1}e^{-t} + c_{2}e^{2t}[/tex]is a solution to the differential equation-----------------------------------------------------------The solution to the initial value problem is:[tex]x(t) = \frac{20}{3}e^{-t} +  \frac{10}{3}e^{2t}[/tex]Step-by-step explanation:We have the following differential equation:[tex]x'' - x' - 2x = 0[/tex]The first step is finding the eigenvalues for this differential equation, that is, finding the roots of the following second order equation:[tex]r^{2} - r - 2 = 0[/tex][tex]\bigtriangleup = (-1)^{2} -4*1*(-2) = 1 + 8 = 9[/tex][tex]r_{1} = \frac{-(-1) + \sqrt{\bigtriangleup}}{2*1} = \frac{1 + 3}{2} = 2[/tex][tex]r_{2} = \frac{-(-1) - \sqrt{\bigtriangleup}}{2*1} = \frac{1 - 3}{2} = -1[/tex]Since 2 and -1 are eigenvalues of the differential equation,[tex]x(t) = c_{1}e^{-t} + c_{2}e^{2t}[/tex]is a solution to the differential equation.Solution of the initial value problem:[tex]x(t) = c_{1}e^{-t} + c_{2}e^{2t}[/tex][tex]x(0) = 10[/tex][tex]10 = c_{1}e^{-0} + c_{2}e^{2*0}[/tex][tex]c_{1} + c_{2} = 10[/tex]---------------------[tex]x'(t) = -c_{1}e^{-t} + 2c_{2}e^{2t}[/tex][tex]x'(0) = 0[/tex][tex]0 = -c_{1}e^{-0} + 2c_{2}e^{2*0}[/tex][tex]-c_{1} + 2c_{2} = 0[/tex][tex]c_{1} = 2c_{2}[/tex]So, we have to solve the following system:[tex]c_{1} + c_{2} = 10[/tex][tex]c_{1} = 2c_{2}[/tex][tex]2c_{2} + c_{2} = 10[/tex][tex]3c_{2} = 10[/tex][tex]c_{2} = \frac{10}{3}[/tex][tex]c_{1} = 2c_{2} = \frac{20}{3}[/tex]The solution to the initial value problem is:[tex]x(t) = \frac{20}{3}e^{-t} +  \frac{10}{3}e^{2t}[/tex]