Y16y64y0 y0 0 y01 y09SolutionIt is a linear homogeneous ode

Y\'\'\'+16y\'\'+64y\'=0, y(0) = 0, y\'(0)=1, y\'\'\'(0)=-9

It is a linear homogeneous ode with constant coefficients

So solution is fo the form

y=exp(kx)

Substituting gives

k^3+16k^2+64k=0

k(K+8)^2=0

k=0,k=-8

k=-8 is a repeated root

y=A+e^{-8x}(B+Cx)

y(0)=A+B=0

So, A=-B

y=-B+e^{-8x}(B+Cx)

y\'(x)=-8e^{-8x}(B+Cx)+e^{-8x}C

y\'(0)=-8B+C=1 , C=1+8B

y\'\'(x)=16e^{-8x}(4Cx+4B-C)

y\'\'(0)=16(4B-C)=-9

16(4B-1-8B)=-9

(4B+1)=9/16

4B=-7/16

B=-7/64

A=7/64

C=1+8B=1-7/8=1/8

C=1/8

$Y\'\'\'+16y\'\'+64y\'=0, y(0) = 0, y\'(0)=1, y\'\'\'(0)=-9SolutionIt is a linear homogeneous ode with constant coefficients So solution is fo the form y=exp(kx$

Y16y64y0 y0 0 y01 y09SolutionIt is a linear homogeneous ode

Solution

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