In each of Problems 11 through 23, use the Laplace transform to solve the given

Question

In each of Problems 11 through 23, use the Laplace transform to solve the given initial value problem. y" - y' - 6y = 0; y(0) = 1, y'(0) = -1 y" + 3y' + 2y = 0; y(0) = 1, y'(0) = 0 y" - 2y' + 2y = 0; y(0) = 0, y'(0) = 1 y" - 4y' + 4y = 0; y(0) = 1, y'(0) = 1 y" - 2y + 4y = 0; y(0) = 2, y'(0) = 0 y" + 2y' + 5y = 0; y(0) = 2, y'(0) = -1 y^(4) - 4y^m + 6y" - 4y' + y = 0; y(0) = 0, y"(0) = 0, y"(0) = 1, y'"(0) = 1 y^(4) - y = 0; y(0) 1, y'(0) = 0, y"(0) = 1, y"'(0) = 0 y^(4) - 4y = 0; y(0) = 1, y'(0) = 0, y"(0) = -2, y'"(0) = 0 y" + omega^2 y = cos 2t, omega^2 notequalto 4; y(0) = 1, y'(0) = 0 y"(0) = -2y' + 2y = cos t; y(0) = 1, y'(0) = 0, y" - 2y' + 2y = e^-t; y(0) = 0, y'(0) = 0 y" - 2y' + 2y = e^-t; y(0) = 0, y'(0) = 1 y" + 2y' + y = 4e^-t; y(0) = 2, y'(0) = -1

Explanation / Answer

First we solve homogeneous ode ie

y''+2y'+y=0

Solution is of the form

y=exp(kt)

Substituting gives

k^2+2k+1=0

So repeated root: k=-1

So, general solution to homogeneous ode is

yh=e^{-t}(A+Bt)

Guess for the particular solution is based on inhomogeneous part ie 4e^{-t}

Usually the guess would be: C e^{-t} and in case e^{-t} is solution to homogeneous ode guess is

C te^{-t} since even that is a solution to homogeneous ode so guess is

yp=Ct^2e^{-t}

Substituting thsi is equation gives

C=2

y=e^{-t}(A+Bt)+2t^2e^{-t}

y(0)=A=2

y'=-y+e^{-t}B+2Cte^{-t}

y'(0)=-y(0)+B=-2+B=-1

B=1

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