Use the Laplace transform to solve the initial value problem y'' - y = 4*delta(t - 2) + t^2 with y(0) = 0 and y'(0) = 2, where delta is the Dirac delta function.

Step-by-step solution

1. 1

   Choose the Laplace transform because of the impulse. The forcing term contains $\delta(t-2)$, which is not a function that can be integrated by ordinary methods; undetermined coefficients and variation of parameters both fail. The Laplace transform turns the impulse into the ordinary exponential $e^{-2s}$, so the whole problem becomes algebra.

2. 2

   Transform both sides, folding in the initial conditions. Using $\mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0)$ with $y(0)=0$, $y'(0)=2$, and the standard transforms $\mathcal{L}\{\delta(t-a)\} = e^{-as}$ and $\mathcal{L}\{t^2\} = 2/s^3$:

   $s^2Y - 2 - Y = 4e^{-2s} + \frac{2}{s^3}$

3. 3

   Solve for $Y(s)$. Factoring $Y$ out of the left side:

   $Y(s) = \frac{2}{s^2-1} + \frac{4e^{-2s}}{s^2-1} + \frac{2}{s^3(s^2-1)}$

   The three pieces are inverted separately — that is the payoff of working in the $s$-domain.

4. 4

   Break the awkward third term with partial fractions. Writing $\dfrac{1}{s^3(s^2-1)} = \dfrac{A}{s} + \dfrac{B}{s^2} + \dfrac{C}{s^3} + \dfrac{Ds+E}{s^2-1}$ and matching coefficients gives $A = -1$, $B = 0$, $C = -1$, $D = 1$, $E = 0$, so

   $\frac{2}{s^3(s^2-1)} = -\frac{2}{s} - \frac{2}{s^3} + \frac{2s}{s^2-1}$

5. 5

   Invert each term. Using $\mathcal{L}^{-1}\{1/(s^2-1)\} = \sinh t$, $\mathcal{L}^{-1}\{s/(s^2-1)\} = \cosh t$, $\mathcal{L}^{-1}\{2/s^3\} = t^2$, and the second shifting theorem $\mathcal{L}^{-1}\{e^{-as}F(s)\} = u(t-a)f(t-a)$:

   $y(t) = 2\sinh t + 2\cosh t - t^2 - 2 + 4u(t-2)\sinh(t-2)$

6. 6

   Simplify and check. Since $2\sinh t + 2\cosh t = 2e^{t}$,

   $y(t) = 2e^{t} - t^2 - 2 + 4u(t-2)\sinh(t-2)$

   Check it: $y(0) = 2 - 0 - 2 = 0$ and $y'(0) = 2 - 0 = 2$, both correct. For $t<2$ the Heaviside term is silent and $y'' - y = (2e^t - 2) - (2e^t - t^2 - 2) = t^2$, as required. At $t=2$ the impulse makes $y'$ jump by exactly $4$ while $y$ itself stays continuous — the physical signature of a unit impulse of strength $4$.