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fiasKO [112]
4 years ago
15

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

Mathematics
1 answer:
Ronch [10]4 years ago
8 0

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)

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