1. **Problem 1: Find the general solution of the differential equation** $$tds = (3t + 1)sdt + t^3 e^{3t} dt$$ **Step 1:** Rewrite the equation in differential form: $$tds - (3t + 1)sdt = t^3 e^{3t} dt$$ Divide both sides by $t$ (assuming $t \neq 0$): $$ds - \left(3 + \frac{1}{t}\right)s dt = t^2 e^{3t} dt$$ This is a linear first-order ODE in $s$: $$\frac{ds}{dt} - \left(3 + \frac{1}{t}\right)s = t^2 e^{3t}$$ **Step 2:** Identify integrating factor $\mu(t)$: $$\mu(t) = e^{\int -\left(3 + \frac{1}{t}\right) dt} = e^{-3t - \ln|t|} = e^{-3t} \cdot \frac{1}{t}$$ **Step 3:** Multiply both sides by $\mu(t)$: $$\frac{d}{dt}\left(s \cdot \frac{e^{-3t}}{t}\right) = t^2 e^{3t} \cdot \frac{e^{-3t}}{t} = t$$ **Step 4:** Integrate both sides: $$s \cdot \frac{e^{-3t}}{t} = \int t dt + C = \frac{t^2}{2} + C$$ **Step 5:** Solve for $s$: $$s = e^{3t} t \left(\frac{t^2}{2} + C\right) = \frac{t^3}{2} e^{3t} + C t e^{3t}$$ --- 2. **Problem 2: Solve the differential equation** $$y' + \frac{y}{x} - \sqrt{y} = 0$$ Rewrite as: $$y' = \sqrt{y} - \frac{y}{x}$$ Let $y = z^2$, so $y' = 2z z'$: $$2z z' = z - \frac{z^2}{x}$$ Divide both sides by $z$ (assuming $z \neq 0$): $$2 z' = 1 - \frac{z}{x}$$ Rewrite: $$2 \frac{dz}{dx} + \frac{z}{x} = 1$$ This is a linear ODE in $z$: $$\frac{dz}{dx} + \frac{z}{2x} = \frac{1}{2}$$ Integrating factor: $$\mu(x) = e^{\int \frac{1}{2x} dx} = e^{\frac{1}{2} \ln|x|} = |x|^{1/2}$$ Multiply both sides: $$\frac{d}{dx} (z x^{1/2}) = \frac{1}{2} x^{1/2}$$ Integrate: $$z x^{1/2} = \int \frac{1}{2} x^{1/2} dx + C = \frac{1}{2} \cdot \frac{2}{3} x^{3/2} + C = \frac{x^{3/2}}{3} + C$$ Solve for $z$: $$z = \frac{x^{3/2}/3 + C}{x^{1/2}} = \frac{x}{3} + C x^{-1/2}$$ Recall $y = z^2$: $$y = \left(\frac{x}{3} + C x^{-1/2}\right)^2$$ --- 3. **Problem 3: Thermometer temperature after 1 minute** Given: - Initial thermometer reading $T_0 = 18^\circ F$ - Room temperature $T_r = 294 K = 294 - 273.15 = 20.85^\circ C$ - After 1 minute, thermometer reads $491^\circ R$ Convert $491^\circ R$ to Celsius: $$T = (491 - 491.67) \times \frac{5}{9} = -0.37^\circ C$$ Newton's law of cooling: $$\frac{dT}{dt} = -k (T - T_r)$$ Solution: $$T(t) = T_r + (T_0 - T_r) e^{-kt}$$ At $t=1$ min: $$-0.37 = 20.85 + (18 - 20.85) e^{-k}$$ Simplify: $$-0.37 - 20.85 = -2.85 e^{-k}$$ $$-21.22 = -2.85 e^{-k}$$ $$e^{-k} = \frac{21.22}{2.85} = 7.45$$ This is impossible since $e^{-k} \leq 1$. Check unit conversions carefully. Convert $18^\circ F$ to Celsius: $$T_0 = (18 - 32) \times \frac{5}{9} = -7.78^\circ C$$ Convert $491^\circ R$ to Celsius: $$T = (491 - 491.67) \times \frac{5}{9} = -0.37^\circ C$$ Now: $$-0.37 = 20.85 + (-7.78 - 20.85) e^{-k}$$ $$-0.37 - 20.85 = -28.63 e^{-k}$$ $$-21.22 = -28.63 e^{-k}$$ $$e^{-k} = \frac{21.22}{28.63} = 0.741$$ Calculate $k$: $$k = -\ln(0.741) = 0.299$$ Find temperature at $t=0$ (already known) and at $t=0$ (initial), so find temperature at $t=0$ (already known). The question asks for temperature after first brought into the room, which is $t=0$, so answer is $-7.78^\circ C$. --- 4. **Problem 4: Radium decay half-life** Given: - Decay rate proportional to amount: $$\frac{dQ}{dt} = -kQ$$ - After 25 years, 1.1% decomposed, so remaining is 98.9%: $$Q(25) = 0.989 Q_0$$ Solution: $$Q(t) = Q_0 e^{-kt}$$ At $t=25$: $$0.989 Q_0 = Q_0 e^{-25k} \Rightarrow e^{-25k} = 0.989$$ Take natural log: $$-25k = \ln(0.989) = -0.01106$$ $$k = \frac{0.01106}{25} = 0.0004424$$ Find half-life $t_{1/2}$: $$Q(t_{1/2}) = \frac{Q_0}{2} = Q_0 e^{-k t_{1/2}}$$ $$\frac{1}{2} = e^{-k t_{1/2}}$$ $$-k t_{1/2} = \ln \frac{1}{2} = -0.6931$$ $$t_{1/2} = \frac{0.6931}{k} = \frac{0.6931}{0.0004424} \approx 1566 \text{ years}$$