Solve y y 2y 0 y0 7 y0 2 by converting to a 1st order s

Solve y\" - y\' - 2y = 0, y(0) = 7, y\'(0) = 2 by converting to a 1^st order system and using the elimination method. No credit for using Laplace transforms.

Solution

y\'=w

y\'\'=w\'

So,

w\'-w-2y=0

w\'=w+2y

y\'=w

y(0)=7,w(0)=2

y=(w\'-w)/2

y\'=(w\'\'-w\')/2

Using y\'=w from second equation gives

2w=w\'\'-w\'

w\'\'-w\'-2w=0

Linear homogeneous odes with constant coefficients so, w=exp(kt)

Substituting gives

k^2-k-2=0

k^2-2k+k-2=-

k=2,-1

w=Ae^{2t}+Be^{-t}

w(0)=2=A+B

y=(w\'-w)/2=(2Ae^{2t}-Be^{-t}-Ae^{2t}-Be^{-t})/2=(Ae^{2t}-2B e^{-t})/2

y=(Ae^{2t}-2B e^{-t})/2

y(0)=7=(A-2B)/2

A-2B=14

A+B=2

B=-4

A=6

y=(6e^{2t}+8 e^{-t})/2=4e^{-t}+3e^{2t}