ODE6.1

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$y''-y=e^{-t},\,\,\,y(0)=1, y'(0)=0\,$

$\mathcal{L}[y''-y] = \mathcal{L}[e^{-t}]\,$

Let $Y = \mathcal{L}[y]\,$

$\mathcal{L}[y''-y] = \mathcal{L}[y'']-\mathcal{L}[y] = s^2 Y -s f(0) - f'(0) - Y\,$

$= s^2 Y - s - Y = \frac{1}{s+1}\,$

At this point all the information from the IC's are encoded into the transform. Solve for $Y$ .

$Y = \frac{s^2+s+1}{(s+1)(s^2-1)}\,$

Find the partial fraction decomposition of $\frac{s^2+s+1}{(s+1)(s^2-1)}\,$

$\frac{s^2+s+1}{(s+1)(s^2-1)} = \frac{s^2+s+1}{(s+1)^2(s-1)} = \frac{A}{(s+1)^2} + \frac{B}{s+1} + \frac{C}{s-1}\,$

Multiply both sides by the common denominator.

$s^2+s+1 = A(s-1) + B(s+1)(s-1) + C(s+1)^2\,$

$s^2+s+1 = s^2(B+C) + s(A+2C) + (-A-B+C)\,$

Compare coefficients.

$B+C=1, \,\,\, A+2C=1, \,\,\, -A-B+C=1\,$

Solve this system of equations. The augmented matrix has equations as rows and variables as columns. The last column is the value of each equation.

$\begin{bmatrix} A & B & C & {}\\ 0 & 1 & 1 & 1 \\ 1 & 0 & 2 & 1 \\ -1&-1 & 1 & 1 \\ \end{bmatrix} = \begin{bmatrix} A & B & C & {}\\ 1 & 0 & 2 & 1 \\ -1&-1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ \end{bmatrix} = \begin{bmatrix} A & B & C & {}\\ 1 & 0 & 2 & 1 \\ 0 &-1 & 3 & 2 \\ 0 & 1 & 1 & 1 \\ \end{bmatrix} = \begin{bmatrix} A & B & C & {}\\ 1 & 0 & 2 & 1 \\ 0 &-1 & 3 & 2 \\ 0 & 0 & 4 & 3 \\ \end{bmatrix} \,$