Question

y" +2y' +17y=0; y(0)=3, y'(0)=17

Answer 1

Answer:

The solution is $y(t)=e^{-t}(\cos 32t + (\frac{5}{8}) \sin 32t)$

Step-by-step explanation:

We need to find the solution of $y''+2y'+17y=0$ with

condition $y(0)=3,\ y'(0)=17$

This is a homogeneous equation with characteristic polynomial

$r^{2}+2r+17=0$

using quadratic formula $x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

$r=\frac{-2\pm \sqrt{2^{2}-4(1)(17)}}{2(1)}$

$r=\frac{-2\pm \sqrt{4-68}}{2}$

$r=\frac{-2\pm \sqrt{-64}}{2}$

$r=\frac{-2\pm 64i}{2}$

$r=-1 \pm 32i$

The general solution for eigen value $a \pm ib$ is

$y(t)=e^{at}(A \cos bt + B \sin bt)$

$y(t)=e^{-t}(A \cos 32t + B \sin 32t)$

Differentiate above with respect to 't'

$y'(t)=-e^{-t}(A \cos 32t + B \sin 32t) + e^{-t}(-32A \sin 32t + 32B \cos 32t)$

Since, y(0)=3

$y(0)=e^{0}(A \cos(0) + B \sin(0))$

$3=(A \cos(0) +0)$

so, A=1

Since, y'(0)=17

$y'(0)=-e^{0}(3 \cos(0) + B \sin(0)) + e^{0}(-32(3) \sin(0) + 32B \cos (0))$

$17=-(3 \cos(0)) + (0 + 32B \cos (0))$

$17=-3 + 32B$

add both the sides by 3,

$17+3 = 32B$

$20= 32B$

divide both the sides, by 32,

$\frac{20}{32}= B$

$\frac{5}{8}= B$

Put the value of constants in $y(t)=e^{-t}(A \cos 32t + B \sin 32t)$

$y(t)=e^{-t}((1) \cos 32t + (\frac{5}{8}) \sin 32t)$

Therefore, the solution is $y(t)=e^{-t}(\cos 32t + (\frac{5}{8}) \sin 32t)$