Question Number 100243 by mhmd last updated on 25/Jun/20

find laplace transforme of the function   f(t)=(a−bt)^2 +cos^2 (wt)?    help me sir ?

Answered by mathmax by abdo last updated on 25/Jun/20

for that let find L(1)  ,L(t^n ) and L(cos(wt)) and use linearity of L  L(1) =∫_0 ^∞  e^(−tx) dx =[−(1/t)e^(−xt) ]_0 ^∞  =(1/t)  L(t^n ) =∫_0 ^∞  x^n  e^(−tx)  dx  =_(tx=u)   ∫_0 ^∞  (u^n /t^n )e^(−u)  (du/t) =(1/t^(n+1) ) ∫_0 ^∞  u^n  e^(−u)  du  =((Γ(n+1))/t^(n+1) ) =((n!)/t^(n+1) )  L(cos(wt)) =∫_0 ^∞  cos(wx)e^(−tx)  dx =Re(∫_0 ^∞  e^(iwx+tx) dx) and   ∫_0 ^∞  e^((−t+iw)x)  dx =[(1/(−t+iw))e^((−t+iw)x) ]_(x=0) ^∞  =−(1/(−t+iw)) =(1/(t−iw)) =((t+iw)/(t^2  +w^2 )) ⇒  L(cos(wt)) =(t/(t^2  +w^2 ))  we have f(t) =a^2 −2abt +b^2  t^2  +((1+cos(2wt))/2) ⇒  L(f(t)) =a^2  L(1)−2ab L(t)+b^2  L(t^2  )+(1/2)L(1)+(1/2)L(cos(2wt))  =(a^2 /t)−((2ab)/t^2 ) +b^2 (2/t^3 ) +(1/(2t)) +(1/2)×(t/(t^2 +4w^2 )) ⇒  L(f(t)) =(a^2 /t)−((2ab)/t^2 ) +((2b^2 )/t^3 ) +(1/(4t)) +(t/(2(t^2  +4w^2 )))

Commented bymathmax by abdo last updated on 25/Jun/20

L(f(t)) =(a^2 /t)−((2ab)/t^2 ) +((2b^2 )/t^3 ) +(1/(2t)) +(t/(2(t^2  +4w^2 )))

Answered by smridha last updated on 25/Jun/20

f(s)=L[(a^2 −2abt+b^2 t^2 )]+(1/2)L[(1+cos2wt)]           =a^2 L[1]−2abL[t]+b^2 L[t^2 ]+(1/2)L[1]+(1/2)L[cos2wt]          =(a^2 /s)−2ab.(1/s^2 )+b^2 .(2/s^3 )+(1/(2s))+(1/2).(s/((s^2 +4w^2 ))).