Inverse Laplace Transform : F(s)= e^(-3s)/((s+1)(s-1))

Related BrainMass Solutions

• Inverse Laplace Transform and Heaviside Step Function : F(s)= e^(-3s)/((s+1)(s-1))

  44928 Inverse Laplace Transform and Heaviside Step Function : F(s)= e^(-3s)/((s+1)(s-1)) Determine the inverse laplace transform of F(s)= e^(-3s)/((s+1)(s-1)) Please see the attached file for the complete solution. Thanks for using BrainMass.

• Inverse Laplace Transform

  212084 Inverse Laplace Transform Please see the attached file for the fully formatted problem. Find the inverse Laplace transform of the function: F(s) = 2e^(-3s)/s - e^(-s)/s You may use a table of Laplace transforms.

• 5 problems regarding Laplace transform

  Compute the inverse transform of (3s^2 + 4s + 1) / (s^4 + 3s^3 +3s^2+2s) 4. Use Laplace transform to calculate the solution to the ODE y"+6y'+8y=u(t) y(0)=0 y'(0)=1 5.

• Laplace transform and differential equations.

  Find the inverse Laplace transform of (s^3+s^2+2/s) / [s^2(s^2+3s+2)] Using this (or otherwise), Find the solution of the equation y"+3y'+2y = 1-t Find the transform of the following functions: f(t) = (1+t^2)[u(t-1)-u(t-2)] where u(t) is the

• Laplace transforms, Inverse Laplace transforms, & Diff Eqns

  Compute the inverse Laplace transform of: X(s) = (3s^2+2s+1) / (s^3+5s^2+8s+4) 3. Use Laplace transforms to compute the solution to the differential equation given in the attachment. Please refer to the attachment.

• Question about Inverse Laplace transforms

  I have taken a simpler inverse transform F(s) = e^(-bs)/s^2 and separated it into two separate laplace transforms F(s) = e^(-bs)/s x 1/s the inverse transform of these individually are u_c(t) and 1 respectively which would make the inverse

• Laplace Transformations and Inverse Laplace Transformations

  Therefore we cannot determine inverse laplace transform of s/s+2 5. Find the inverse Laplace transform of 1/s+5. Since L(e^-5t) = 1/s + 5 Then L^-1 (1/s + 5) = e^-5t 6. Find the inverse Laplace transform of 1/( s + 5 ) ( s 2 + 1).

• Laplace transformations

  if f(t) <-----> F(s), then f(t-a)U(t-a) <-----> F(s)e^-(as) (d)L^-1 [ 1/{S^2(S^2+1) ] = L^-1 [ (1/S^2) - {1/(S^2+1)} ] = t - Sin t (e) L^-1 [ 3s/(s^2+4s+8 ] = L^-1 [ 3s/{(s+2)^2+2^2 }] = 3e^(-2t)[ Cos 2t - Sin 2t ]

• How do you find the inverse laplace transform of L-1[(e^-2s)/(s^2)]?

  21207 How do you find the inverse laplace transform? Find the inverse Laplace transform: (L-1) * [(e^-2s) / (s^2)] I will use the notation "^" for the exponent operator and "^-1" for the Inverse function.