1. **Problem statement:** We are given two piecewise functions to analyze and understand: - Problem 65: $$f(t) = \begin{cases} t, & 0 \leq t < 1 \\ 0, & 1 \leq t < 2 \\ \sin(\pi t), & t \geq 2 \end{cases}$$ - Problem 75: Solve the differential equation $$y'' + 4y = \sin t \cdot U(t - 2\pi), \quad y(0) = 1, \quad y'(0) = 0,$$ where $U(t - 2\pi)$ is the Heaviside step function activating at $t=2\pi$. --- 2. **Understanding Problem 65:** This is a piecewise function defined in three intervals: - For $0 \leq t < 1$, $f(t) = t$ (a linear function). - For $1 \leq t < 2$, $f(t) = 0$ (zero function). - For $t \geq 2$, $f(t) = \sin(\pi t)$ (a sinusoidal function). This function is continuous at the boundaries if values match: - At $t=1$, from left $f(1^-) = 1$, from right $f(1) = 0$, so there is a jump discontinuity. - At $t=2$, from left $f(2^-) = 0$, from right $f(2) = \sin(2\pi) = 0$, continuous. --- 3. **Understanding Problem 75:** We have a second order linear ODE with forcing term $f(t) = \sin t \cdot U(t - 2\pi)$. - The Heaviside function $U(t - 2\pi)$ is 0 for $t < 2\pi$ and 1 for $t \geq 2\pi$. - So the forcing term is zero until $t=2\pi$, then it becomes $\sin t$. Initial conditions are $y(0) = 1$ and $y'(0) = 0$. --- 4. **Solving Problem 65:** Since the problem only asks to analyze the function, we describe it and its behavior: - For $0 \leq t < 1$, $f(t) = t$ is increasing linearly from 0 to 1. - For $1 \leq t < 2$, $f(t) = 0$ is flat at zero. - For $t \geq 2$, $f(t) = \sin(\pi t)$ oscillates with period $2$ (since period of $\sin(\pi t)$ is $\frac{2\pi}{\pi} = 2$). This function is piecewise continuous with a jump at $t=1$. --- 5. **Solving Problem 75:** Step 1: Solve the homogeneous equation $$y'' + 4y = 0.$$ Characteristic equation: $$r^2 + 4 = 0 \implies r = \pm 2i.$$ General solution: $$y_h = C_1 \cos 2t + C_2 \sin 2t.$$ Step 2: Use Laplace transform to solve with forcing term involving Heaviside function. Let $Y(s) = \mathcal{L}\{y(t)\}$. Laplace transform of $y'' + 4y$: $$s^2 Y(s) - s y(0) - y'(0) + 4 Y(s) = \mathcal{L}\{\sin t \cdot U(t - 2\pi)\}.$$ Given $y(0) = 1$, $y'(0) = 0$: $$s^2 Y(s) - s \cdot 1 + 4 Y(s) = e^{-2\pi s} \mathcal{L}\{\sin(t + 2\pi)\}.$$ Using the second shifting theorem: $$\mathcal{L}\{f(t - a) U(t - a)\} = e^{-as} F(s).$$ Rewrite forcing term: $$\sin t \cdot U(t - 2\pi) = \sin((t - 2\pi) + 2\pi) U(t - 2\pi) = \sin(t - 2\pi) U(t - 2\pi)$$ (since $\sin$ is $2\pi$ periodic). So, $$\mathcal{L}\{\sin t \cdot U(t - 2\pi)\} = e^{-2\pi s} \mathcal{L}\{\sin t\} = e^{-2\pi s} \frac{1}{s^2 + 1}.$$ Step 3: Substitute into equation: $$s^2 Y(s) - s + 4 Y(s) = e^{-2\pi s} \frac{1}{s^2 + 1}.$$ Group terms: $$(s^2 + 4) Y(s) = s + e^{-2\pi s} \frac{1}{s^2 + 1}.$$ Step 4: Solve for $Y(s)$: $$Y(s) = \frac{s}{s^2 + 4} + e^{-2\pi s} \frac{1}{(s^2 + 4)(s^2 + 1)}.$$ Step 5: Inverse Laplace transform: - First term: $$\mathcal{L}^{-1}\left\{\frac{s}{s^2 + 4}\right\} = \cos 2t.$$ - Second term involves shift: $$\mathcal{L}^{-1}\left\{e^{-2\pi s} F(s)\right\} = U(t - 2\pi) f(t - 2\pi),$$ where $$F(s) = \frac{1}{(s^2 + 4)(s^2 + 1)}.$$ Step 6: Find $f(t) = \mathcal{L}^{-1}\{F(s)\}$. Use partial fractions: $$\frac{1}{(s^2 + 1)(s^2 + 4)} = \frac{A s + B}{s^2 + 1} + \frac{C s + D}{s^2 + 4}.$$ Since numerator is constant 1, and denominators are quadratic, try: $$\frac{1}{(s^2 + 1)(s^2 + 4)} = \frac{E}{s^2 + 1} + \frac{F}{s^2 + 4}.$$ Multiply both sides by denominator: $$1 = E (s^2 + 4) + F (s^2 + 1) = (E + F) s^2 + 4E + F.$$ Equate coefficients: - Coefficient of $s^2$: $0 = E + F$ - Constant term: $1 = 4E + F$ From first: $F = -E$ Substitute into second: $$1 = 4E - E = 3E \implies E = \frac{1}{3}, \quad F = -\frac{1}{3}.$$ So, $$F(s) = \frac{1/3}{s^2 + 1} - \frac{1/3}{s^2 + 4}.$$ Step 7: Inverse Laplace transforms: - $$\mathcal{L}^{-1}\left\{\frac{1}{s^2 + a^2}\right\} = \frac{\sin a t}{a}.$$ Therefore, $$f(t) = \frac{1}{3} \sin t - \frac{1}{3} \frac{\sin 2t}{2} = \frac{1}{3} \sin t - \frac{1}{6} \sin 2t.$$ Step 8: Final solution: $$y(t) = \cos 2t + U(t - 2\pi) \left( \frac{1}{3} \sin (t - 2\pi) - \frac{1}{6} \sin 2(t - 2\pi) \right).$$ Since sine is $2\pi$ periodic, $$\sin (t - 2\pi) = \sin t, \quad \sin 2(t - 2\pi) = \sin 2t,$$ so $$y(t) = \cos 2t + U(t - 2\pi) \left( \frac{1}{3} \sin t - \frac{1}{6} \sin 2t \right).$$ --- **Final answers:** - Problem 65: The piecewise function is $$f(t) = \begin{cases} t, & 0 \leq t < 1 \\ 0, & 1 \leq t < 2 \\ \sin(\pi t), & t \geq 2 \end{cases}$$ with a jump discontinuity at $t=1$ and sinusoidal behavior for $t \geq 2$. - Problem 75: The solution to the ODE is $$y(t) = \cos 2t + U(t - 2\pi) \left( \frac{1}{3} \sin t - \frac{1}{6} \sin 2t \right).$$