1. Statement of the problem. Problem: Solve the ordinary differential equation using the Laplace transform method. $$y'' - 6y' + 16y = 2e^{2t},\quad y(0) = -1,\ y'(0) = -4$$ 2. Formula and important rules used. Use the Laplace transform of derivatives and the transform of an exponential. The key formulas are $\mathcal{L}\{y''\}=s^2Y - s y(0) - y'(0)$, $\mathcal{L}\{y'\}=sY - y(0)$, and $\mathcal{L}\{e^{at}\}=\frac{1}{s-a}$. Remember to solve algebraically for $Y(s)$ and then use partial fractions and inverse transforms. 3. Apply the Laplace transform to both sides and substitute initial conditions. Taking transforms gives $$ (s^2Y - s y(0) - y'(0)) - 6(sY - y(0)) + 16Y = \frac{2}{s-2}$$ Substitute $y(0)=-1$ and $y'(0)=-4$ to obtain $$ (s^2Y + s + 4) - 6(sY + 1) + 16Y = \frac{2}{s-2}$$ Collect $Y$ terms and constants to get $$Y(s^2 - 6s + 16) + (s - 2) = \frac{2}{s-2}$$ Move the non-$Y$ term to the right-hand side to isolate the $Y$ factor. $$Y(s^2 - 6s + 16) = \frac{2}{s-2} - (s-2)$$ Combine the right-hand side over a common denominator. $$Y(s^2 - 6s + 16) = \frac{2 - (s-2)^2}{s-2}$$ Expand the numerator. $$Y(s^2 - 6s + 16) = \frac{-s^2 + 4s - 2}{s-2}$$ Now divide both sides by the characteristic polynomial and show cancellation of the factor on the left. $$\frac{Y(\cancel{s^2 - 6s + 16})}{\cancel{s^2 - 6s + 16}} = \frac{-s^2 + 4s - 2}{(s-2)(s^2 - 6s + 16)}$$ Therefore $$Y = \frac{-s^2 + 4s - 2}{(s-2)(s^2 - 6s + 16)} = -\frac{s^2 - 4s + 2}{(s-2)(s^2 - 6s + 16)}$$ 4. Partial fraction decomposition. Assume a decomposition of the form $$-\frac{s^2 - 4s + 2}{(s-2)(s^2 - 6s + 16)} = \frac{A}{s-2} + \frac{Bs + C}{s^2 - 6s + 16}$$ Multiply both sides by the denominator and equate coefficients to obtain $$-s^2 + 4s - 2 = A(s^2 - 6s + 16) + (Bs + C)(s-2).$$ Expand the right-hand side and collect like terms. This gives the system from coefficients of $s^2$, $s$, and constant: $A + B = -1$. $-6A - 2B + C = 4$. $16A - 2C = -2$. Solve the linear system step by step. From $A + B = -1$ we have $B = -1 - A$. From $16A - 2C = -2$ we get $8A - C = -1$ so $C = 8A + 1$. Substitute $B$ and $C$ into the $s$-coefficient equation and solve: $-6A - 2(-1 - A) + (8A + 1) = 4$. Simplify to find $4A + 3 = 4$ so $A = \frac{1}{4}$. Then $B = -1 - \frac{1}{4} = -\frac{5}{4}$ and $C = 8\cdot\frac{1}{4} + 1 = 3$. Thus the partial fractions are $$Y = \frac{1}{4}\cdot\frac{1}{s-2} + \frac{-\tfrac{5}{4}s + 3}{s^2 - 6s + 16}.$$ 5. Prepare terms for inverse Laplace by completing the square. Write the quadratic as $s^2 - 6s + 16 = (s-3)^2 + 7$. Rewrite the numerator of the second fraction in terms of $(s-3)$: $$-\frac{5}{4}s + 3 = -\frac{5}{4}(s-3) - \frac{3}{4}.$$ So $$Y = \frac{1}{4}\frac{1}{s-2} - \frac{5}{4}\frac{s-3}{(s-3)^2 + 7} - \frac{3}{4}\frac{1}{(s-3)^2 + 7}.$$ Recall the inverse transforms $\mathcal{L}^{-1}\{\frac{s-a}{(s-a)^2 + b^2}\} = e^{at}\cos(bt)$ and $\mathcal{L}^{-1}\{\frac{b}{(s-a)^2 + b^2}\} = e^{at}\sin(bt)$. 6. Invert each term to obtain the solution $y(t)$. The first term gives $\frac{1}{4}e^{2t}$. The second term gives $-\frac{5}{4}e^{3t}\cos(\sqrt{7}t)$. For the third term write $-\frac{3}{4}\frac{1}{(s-3)^2 + 7} = -\frac{3}{4\sqrt{7}}\cdot\frac{\sqrt{7}}{(s-3)^2 + (\sqrt{7})^2}$, whose inverse is $-\frac{3}{4\sqrt{7}}e^{3t}\sin(\sqrt{7}t)$. 7. Final answer. Combining all terms, the solution satisfying the initial conditions is $$y(t) = \frac{1}{4}e^{2t} - \frac{5}{4}e^{3t}\cos\left(\sqrt{7}\,t\right) - \frac{3}{4\sqrt{7}}e^{3t}\sin\left(\sqrt{7}\,t\right).$$ This function is the inverse Laplace transform of $Y(s)$ and satisfies $y(0)=-1$ and $y'(0)=-4$.