Question

Y''-6y'+9y=t y(0)=0 y'(0)=1

Answer 1

L[ y(t) ] = Y(s) 

L[ y' ] = sY-y(0) = sY-0 = sY 

L[ y'' ] = s^2*Y - sy(0) - y'(0) = s^2*Y - 1 

laplace trasform both sides 

L[ y'' - 6y' + 9y ] = L[ t ] 

= L[ y'' ] - 6 L[ y' ] + 9 L[ y ] = 1/s 

= [ s^2*Y - 1 ] - 6[ sY ] + 9Y = 1/s 

( s^2 - 6s + 9 ) Y - 1 = 1/s 

⇒ ( s^2 - 6s + 9 ) Y = (1/s) + 1 

⇒Y = [ 1 / ( s(s^2 - 6s + 9 ) ) ] + [ 1 / ( s^2 - 6s + 9 ) ] 

Let inverse laplace trasform , find y(t) : 

y(t) = L^(-1)[ Y(s) ] = L^(-1) { [ 1 / ( s(s^2 - 6s + 9 ) ) ] + [ 1 / ( s^2 - 6s + 9 ) ] } 

= [ (1/3)*t*e^(3t) - (1/9)*e^(3t) + (1/9) ] + [ t*e^(3t) ] 

= (4/3)*t*e^(3t) - (1/9)*e^(3t) + (1/9)

Answer 2

Hello,

I am going to try.

A particular solution is y=pt+q ,y'=p, y"=0
t=y"-6y'+9y=0-6p+9(pt+q)==>9pt+9q-6p=t
==>9p=1 ==>p=1/9
and 9q-6p=0==>-6/9+9q=0==>q=2/27



General solution of the homogene equation

y"-6y'+9y=0
Δ=6²-4*9=0
y=6/2=3
Solution is y=(at+b)e^(3t)

Solution: y=(at+b)e^(3t)+t/9+2/27
y(0)=0 ==>0=b+2/27==>b=-2/27

y'(t)=e^(3t)*3*(at+b)+a e^(3t)
y'(0)=0==>0=3(-2/27)+a==>a=2/9

Solution: y=(2/9t-2/27)e^(3t)+t/9 +2/27