Subjects differential equations

Laplace Ivp Fee5Ff

Step-by-step solutions with LaTeX - clean, fast, and student-friendly.

Use the AI math solver

1. **State the problem:** Solve the initial value problem $$y'' - 6y' + 5y = 3e^{2t}$$ with initial conditions $$y(0) = 2$$ and $$y'(0) = 3$$ using the Laplace transform method. 2. **Recall the Laplace transform formulas:** - $$\mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0)$$ - $$\mathcal{L}\{y'\} = sY(s) - y(0)$$ - $$\mathcal{L}\{y\} = Y(s)$$ - $$\mathcal{L}\{e^{at}\} = \frac{1}{s - a}$$ for $$s > a$$ 3. **Apply the Laplace transform to both sides:** $$\mathcal{L}\{y'' - 6y' + 5y\} = \mathcal{L}\{3e^{2t}\}$$ Substitute transforms: $$s^2Y(s) - sy(0) - y'(0) - 6(sY(s) - y(0)) + 5Y(s) = \frac{3}{s - 2}$$ 4. **Plug in initial values:** $$s^2Y(s) - 2s - 3 - 6(sY(s) - 2) + 5Y(s) = \frac{3}{s - 2}$$ 5. **Expand and group terms:** $$s^2Y(s) - 2s - 3 - 6sY(s) + 12 + 5Y(s) = \frac{3}{s - 2}$$ $$\Rightarrow (s^2 - 6s + 5)Y(s) + (-2s - 3 + 12) = \frac{3}{s - 2}$$ Simplify constants: $$-2s + 9$$ So: $$ (s^2 - 6s + 5)Y(s) - 2s + 9 = \frac{3}{s - 2}$$ 6. **Isolate $$Y(s)$$:** $$Y(s) = \frac{3}{(s - 2)(s^2 - 6s + 5)} + \frac{2s - 9}{s^2 - 6s + 5}$$ 7. **Factor the quadratic:** $$s^2 - 6s + 5 = (s - 1)(s - 5)$$ Rewrite: $$Y(s) = \frac{3}{(s - 2)(s - 1)(s - 5)} + \frac{2s - 9}{(s - 1)(s - 5)}$$ 8. **Partial fraction decomposition:** Decompose $$\frac{3}{(s - 2)(s - 1)(s - 5)} = \frac{A}{s - 2} + \frac{B}{s - 1} + \frac{C}{s - 5}$$ Multiply both sides by denominator: $$3 = A(s - 1)(s - 5) + B(s - 2)(s - 5) + C(s - 2)(s - 1)$$ Plug in $$s=2$$: $$3 = A(2 - 1)(2 - 5) = A(1)(-3) = -3A \Rightarrow A = -1$$ Plug in $$s=1$$: $$3 = B(1 - 2)(1 - 5) = B(-1)(-4) = 4B \Rightarrow B = \frac{3}{4}$$ Plug in $$s=5$$: $$3 = C(5 - 2)(5 - 1) = C(3)(4) = 12C \Rightarrow C = \frac{1}{4}$$ So: $$\frac{3}{(s - 2)(s - 1)(s - 5)} = \frac{-1}{s - 2} + \frac{3/4}{s - 1} + \frac{1/4}{s - 5}$$ 9. **Decompose $$\frac{2s - 9}{(s - 1)(s - 5)} = \frac{D}{s - 1} + \frac{E}{s - 5}$$** Multiply both sides: $$2s - 9 = D(s - 5) + E(s - 1)$$ Plug in $$s=1$$: $$2(1) - 9 = D(1 - 5) + E(0) \Rightarrow -7 = -4D \Rightarrow D = \frac{7}{4}$$ Plug in $$s=5$$: $$2(5) - 9 = D(0) + E(5 - 1) \Rightarrow 1 = 4E \Rightarrow E = \frac{1}{4}$$ 10. **Rewrite $$Y(s)$$:** $$Y(s) = \left(-\frac{1}{s - 2} + \frac{3/4}{s - 1} + \frac{1/4}{s - 5}\right) + \left(\frac{7/4}{s - 1} + \frac{1/4}{s - 5}\right)$$ Combine like terms: $$Y(s) = -\frac{1}{s - 2} + \left(\frac{3/4} + \frac{7/4}\right)\frac{1}{s - 1} + \left(\frac{1/4} + \frac{1/4}\right)\frac{1}{s - 5}$$ $$Y(s) = -\frac{1}{s - 2} + \frac{10/4}{s - 1} + \frac{2/4}{s - 5} = -\frac{1}{s - 2} + \frac{5/2}{s - 1} + \frac{1/2}{s - 5}$$ 11. **Take inverse Laplace transform:** $$y(t) = -e^{2t} + \frac{5}{2}e^{t} + \frac{1}{2}e^{5t}$$ **Final solution:** $$\boxed{y(t) = -e^{2t} + \frac{5}{2}e^{t} + \frac{1}{2}e^{5t}}$$