y''-3y'+2y=1+cost+e^-t

Expert Solution

Colby Messinger answered 1 year ago

use laplace to answer y"-3y'+2y=1+cost+e^-t,y(0)=1,y'(0)=0

Solve the ODE y"+3y'+2y=(cosx)+(x^2)+(e^-1)

y''-3y+2y=e^3t y(0)=1 y'(0)=0 laplace transformation

y"-3y'+2y=4t-8 , y(0)=2 , y'(0)=7 y(t)=?

y''+ 3y'+2y=e^t y(0)=1 y'(0)=-6 Solve using Laplace transforms. Then, solve using undetermined coefficients. Then, solve using...

y''+ 3y'+2y=e^t y(0)=1 y'(0)=-6 Solve using Laplace transforms. Then, solve using undetermined coefficients. Then, solve using variation of parameters.

Solve y''-y'-2y=e^t using variation of parameters.

Solve the equation below for y(t): y''+2y'-3y=8u(t-3): y(0) = 0; y'(0)=0

Solve the differential equation using the Laplace transform. y''' + 3y''+2y' = 100e-t , y(0) =...

Solve the differential equation using the Laplace transform. y''' + 3y''+2y' = 100e-t , y(0) = 0, y'(0) = 0, y''(0) = 0

Solve the differential equation by Laplace transform y^(,,) (t)-2y^' (t)-3y(t)=sint where y^' (0)=0 ,y=(0)=0

Solve by variation of parameters: A. y"−9y = 1/(1 − e^(3t)) B. y" +2y'+26y = e^-t/sin(5t)

Question