Menu Close

Use-Laplace-transform-to-solve-the-differential-equation-d-2-v-t-dt-2-6-dv-t-dt-8v-t-2u-t-when-v-0-1-and-v-0-2-




Question Number 184936 by MikeH last updated on 14/Jan/23
Use Laplace transform to solve the differential  equation   ((d^2 v(t))/dt^2 ) +6((dv(t))/dt) + 8v(t) = 2u(t)    when v(0) = 1 and v^β€’ (0) = βˆ’2
UseLaplacetransformtosolvethedifferentialequationd2v(t)dt2+6dv(t)dt+8v(t)=2u(t)whenv(0)=1andvβˆ™(0)=βˆ’2
Answered by hmr last updated on 15/Jan/23
L{vβ€²β€²} + 6L{vβ€²} + 8L{v} = 2L{u}  β†’   (s^2  L{v} βˆ’ sv(0) βˆ’vβ€²(0))   + 6(sL{v} βˆ’ v(0))  + 8L{v} = 2L{u}  β†’  (s^2  + 6s + 8)L{v}βˆ’s+2βˆ’6 = 2L{u}  β†’  (s^2  + 6s + 8)L{v} = 2L{u} + s + 4  β†’  L{v} = ((2L{u} + s + 4)/(s^2  + 6s + 8))  β†’  L{v} = ((2L{u} + (s + 4))/((s + 4)(s + 2)))  β†’  L{v} =  ((2L{u})/((s + 4)(s + 2))) + (1/(s + 2))  β†’  L{v} =  ((2L{u})/((s + 4)(s + 2))) + L{e^(βˆ’2t) }  β†’  L{v} =  ((4L{u})/(s + 4)) βˆ’ ((2L{u})/(s + 2)) + L{e^(βˆ’2t) }  β†’  L{v} =  4L{e^(βˆ’4t) } L{u} βˆ’ 2L{e^(βˆ’2t) } L{u}+ L{e^(βˆ’2t) }  Sorry thatβ€²s all I know.
L{vβ€³}+6L{vβ€²}+8L{v}=2L{u}β†’(s2L{v}βˆ’sv(0)βˆ’vβ€²(0))+6(sL{v}βˆ’v(0))+8L{v}=2L{u}β†’(s2+6s+8)L{v}βˆ’s+2βˆ’6=2L{u}β†’(s2+6s+8)L{v}=2L{u}+s+4β†’L{v}=2L{u}+s+4s2+6s+8β†’L{v}=2L{u}+(s+4)(s+4)(s+2)β†’L{v}=2L{u}(s+4)(s+2)+1s+2β†’L{v}=2L{u}(s+4)(s+2)+L{eβˆ’2t}β†’L{v}=4L{u}s+4βˆ’2L{u}s+2+L{eβˆ’2t}β†’L{v}=4L{eβˆ’4t}L{u}βˆ’2L{eβˆ’2t}L{u}+L{eβˆ’2t}Sorrythatβ€²sallIknow.
Commented by MikeH last updated on 15/Jan/23
thanks a lot
thanksalot
Commented by hmr last updated on 15/Jan/23
Youβ€²re Welcome
Youβ€²reWelcome
Commented by hmr last updated on 15/Jan/23
I edited the last line, It was wrong!
Ieditedthelastline,Itwaswrong!

Leave a Reply

Your email address will not be published. Required fields are marked *