Answer:The solution is [tex]y(t)=e^{-t}(\cos 32t + (\frac{5}{8}) \sin 32t)[/tex]Step-by-step explanation: We need to find the solution of [tex]y''+2y'+17y=0[/tex] with
condition [tex]y(0)=3,\ y'(0)=17[/tex]
This is a homogeneous equation with characteristic polynomial
[tex]r^{2}+2r+17=0[/tex]
using quadratic formula [tex]x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}[/tex]
[tex]r=\frac{-2\pm \sqrt{2^{2}-4(1)(17)}}{2(1)}[/tex]
[tex]r=\frac{-2\pm \sqrt{4-68}}{2}[/tex]
[tex]r=\frac{-2\pm \sqrt{-64}}{2}[/tex]
[tex]r=\frac{-2\pm 64i}{2}[/tex]
[tex]r=-1 \pm 32i[/tex]The general solution for eigen value [tex]a \pm ib[/tex] is [tex]y(t)=e^{at}(A \cos bt + B \sin bt)[/tex][tex]y(t)=e^{-t}(A \cos 32t + B \sin 32t)[/tex]Differentiate above with respect to 't'[tex]y'(t)=-e^{-t}(A \cos 32t + B \sin 32t) + e^{-t}(-32A \sin 32t + 32B \cos 32t)[/tex]Since, y(0)=3[tex]y(0)=e^{0}(A \cos(0) + B \sin(0))[/tex][tex]3=(A \cos(0) +0)[/tex]so, A=1Since, y'(0)=17[tex]y'(0)=-e^{0}(3 \cos(0) + B \sin(0)) + e^{0}(-32(3) \sin(0) + 32B \cos (0))[/tex][tex]17=-(3 \cos(0)) + (0 + 32B \cos (0))[/tex][tex]17=-3 + 32B[/tex]add both the sides by 3,[tex]17+3 = 32B[/tex][tex]20= 32B[/tex]divide both the sides, by 32,[tex]\frac{20}{32}= B[/tex][tex]\frac{5}{8}= B[/tex]Put the value of constants in [tex]y(t)=e^{-t}(A \cos 32t + B \sin 32t)[/tex][tex]y(t)=e^{-t}((1) \cos 32t + (\frac{5}{8}) \sin 32t)[/tex]Therefore, the solution is [tex]y(t)=e^{-t}(\cos 32t + (\frac{5}{8}) \sin 32t)[/tex]