Particular solution for auxiliary solution

Question

This is the question

I have done the 1a) and b) but i got stucked at c.

For 1a ii) I have got $Ae^{-4x} + Bx^{-4x}$

Answer 1

For ii 1a

$$x''+8x'+16=0$$ $$\implies r^2+8r+16=0 \implies (r+4)^2=0 \implies r=-4$$ With a double root the solution to the differential equation is $$x(t)=Ae^{-4t}+B\color{red}{t}e^{-4t}$$ Now try to find the constants A and B with the initial conditions given $$x(0)=1 \implies x(0)=Ae^{0}+B\times \color{red}{0 \times }e^{0}$$ For the second IC $$x'(0)=1 $$ Differentiate the solution first $$x(t)=Ae^{-4t}+B\color{red}{t}e^{-4t}$$ $$\implies x'(t)=.........$$