Consider the following. x'' + 2x' + x = 0, x(0) = 8, x'(0) = −8; family of solutions x = C1e−t + C2te−t Show that the family of solutions satisfies the equation for all values of the constants. (Enter your answers in terms of t.)

Answer:Everything is verified in the step-by-step explanation.Step-by-step explanation:We have the following differential equation:[tex]x'' + 2x' + x = 0[/tex]This differential equation has the following characteristic polynomial:[tex]r^{2} + 2r + 1 = 0[/tex]This polynomial has two repeated roots of [tex]r = -1[/tex].Since the roots are repeated, our solution has the following format:[tex]x(t) = c_{1}e^{-t} + c_{2}te^{-t}[/tex]This shows that the family of solutions satisfies the equation for all values of the constants. The values of the constants depends on the initial conditions.Lets solve the system with the initial conditions given in the exercise.[tex]x(0) = 8[/tex][tex]c_{1}e^{0} + c_{2}(0)e^{0} = 8[/tex][tex]c_{1} = 8[/tex]--------------------[tex]x'(0) = 8[/tex][tex]x(t) = c_{1}e^{-t} + c_{2}te^{-t}[/tex][tex]x'(t) = -c_{1}e^{-t} + c_{2}e^{-t} - c_{2}te^{-t}[/tex][tex]-c_{1}e^{0} + c_{2}e^{0} - c_{2}(0)e^{0} = 8[/tex][tex]-c_{1} + c_{2} = 8[/tex][tex]c_{2} = 8 + c_{1}[/tex][tex]c_{2} = 8 + 8[/tex][tex]c_{2} = 16[/tex]With these initial conditions, we have the following solution[tex]x(t) = 8e^{-t} + 16te^{-t}[/tex]