Ode Laplace 5A6359

1. **Problem Statement:** We address multiple problems involving second order ordinary differential equations (ODEs) and Laplace transforms. --- 2. **5a: Existence and Uniqueness of IVP for Second Order ODEs** The general form of a second order ODE is: $$y'' = f(t,y,y')$$ with initial conditions: $$y(t_0) = y_0, \quad y'(t_0) = y_0'$$ **Existence and Uniqueness Theorem:** If $f$ and $\frac{\partial f}{\partial y}$, $\frac{\partial f}{\partial y'}$ are continuous in a region containing $(t_0,y_0,y_0')$, then there exists a unique solution $y(t)$ to the IVP on some interval around $t_0$. This means the solution exists and is unique near the initial point. --- 3. **5b: Largest Interval of Validity for** $$(t^2 - 3t)y'' + t y' - (t + 3)y = e^t, \quad y(1) = 2, \quad y'(1) = 1$$ Rewrite in standard form: $$y'' + \frac{t}{t^2 - 3t} y' - \frac{t + 3}{t^2 - 3t} y = \frac{e^t}{t^2 - 3t}$$ The coefficients are undefined where the denominator is zero: $$t^2 - 3t = t(t - 3) = 0 \implies t = 0 \text{ or } t = 3$$ Since initial condition is at $t=1$, the largest interval around 1 avoiding singularities is: $$(0,3)$$ --- 4. **5c: General Solution of** $$y'' - 3y' + 4y = 3e^{2t}$$ **Step 1: Solve homogeneous equation:** $$y'' - 3y' + 4y = 0$$ Characteristic equation: $$r^2 - 3r + 4 = 0$$ Discriminant: $$\Delta = (-3)^2 - 4 \times 1 \times 4 = 9 - 16 = -7 < 0$$ Roots: $$r = \frac{3 \pm \sqrt{-7}}{2} = \frac{3}{2} \pm \frac{\sqrt{7}}{2}i$$ Homogeneous solution: $$y_h = e^{\frac{3}{2}t} \left(C_1 \cos\left(\frac{\sqrt{7}}{2} t\right) + C_2 \sin\left(\frac{\sqrt{7}}{2} t\right)\right)$$ **Step 2: Find particular solution $y_p$ using method of undetermined coefficients:** Guess: $$y_p = Ae^{2t}$$ Compute derivatives: $$y_p' = 2Ae^{2t}, \quad y_p'' = 4Ae^{2t}$$ Substitute into ODE: $$4Ae^{2t} - 3(2Ae^{2t}) + 4(Ae^{2t}) = 3e^{2t}$$ Simplify: $$4A - 6A + 4A = 3 \implies 2A = 3 \implies A = \frac{3}{2}$$ **Step 3: General solution:** $$y = y_h + y_p = e^{\frac{3}{2}t} \left(C_1 \cos\left(\frac{\sqrt{7}}{2} t\right) + C_2 \sin\left(\frac{\sqrt{7}}{2} t\right)\right) + \frac{3}{2} e^{2t}$$ --- 5. **6a: Definition of Laplace Transform** The Laplace transform of a function $f(t)$, defined for $t \geq 0$, is: $$\mathcal{L}\{f(t)\} = F(s) = \int_0^{\infty} e^{-st} f(t) \, dt$$ where $s$ is a complex number parameter. --- 6. **6b: Laplace Transform from First Principles** i. For $f(t) = 1$: $$F(s) = \int_0^{\infty} e^{-st} \cdot 1 \, dt = \left[-\frac{1}{s} e^{-st} \right]_0^{\infty} = \frac{1}{s}, \quad \text{for } \Re(s) > 0$$ ii. For $f(t) = t^n$, $n$ integer $\geq 0$: $$F(s) = \int_0^{\infty} e^{-st} t^n \, dt = \frac{n!}{s^{n+1}}, \quad \Re(s) > 0$$ (This is a standard result derived by integration by parts.) iii. For $f(t) = e^t$: $$F(s) = \int_0^{\infty} e^{-st} e^t \, dt = \int_0^{\infty} e^{-(s-1)t} \, dt = \frac{1}{s-1}, \quad \Re(s) > 1$$ --- 7. **6c: Solve IVP** $$y'' + 2y' + 101y = 0, \quad y(0) = 1, \quad y'(0) = 0$$ **Step 1: Characteristic equation:** $$r^2 + 2r + 101 = 0$$ Discriminant: $$\Delta = 2^2 - 4 \times 1 \times 101 = 4 - 404 = -400 < 0$$ Roots: $$r = \frac{-2 \pm \sqrt{-400}}{2} = -1 \pm 10i$$ **Step 2: General solution:** $$y = e^{-t} (C_1 \cos 10t + C_2 \sin 10t)$$ **Step 3: Apply initial conditions:** At $t=0$: $$y(0) = e^0 (C_1 \cdot 1 + C_2 \cdot 0) = C_1 = 1$$ Derivative: $$y' = e^{-t}(-C_1 \cos 10t - 10 C_1 \sin 10t - C_2 \sin 10t + 10 C_2 \cos 10t)$$ At $t=0$: $$y'(0) = (-C_1 + 10 C_2) = 0 \implies -1 + 10 C_2 = 0 \implies C_2 = \frac{1}{10}$$ **Final solution:** $$y = e^{-t} \left( \cos 10t + \frac{1}{10} \sin 10t \right)$$