1. The order of a differential equation (D.E.) is the highest derivative present. The degree is the power of the highest order derivative if the equation is polynomial in derivatives. 2. Given $\frac{dy}{dx} = x^2$ and the curve passes through $(1,1)$. Integrate: $y = \int x^2 dx = \frac{x^3}{3} + C$. Use the point: $1 = \frac{1^3}{3} + C \Rightarrow C = 1 - \frac{1}{3} = \frac{2}{3}$. So, $y = \frac{x^3}{3} + \frac{2}{3}$. 3. (Problem 3 not fully specified, no equation provided.) 4. Solve $(\cos x \cos y - \cot x) dx - \sin x \sin y dy = 0$. Rearrange to get $M dx + N dy = 0$ where $M = \cos x \cos y - \cot x$, $N = - \sin x \sin y$. Check exactness or use an integrating factor. 5. (Problem 5 not specified.) 6. General solution: $y = C_1 x + C_2 e^x$. Differentiate: $y' = C_1 + C_2 e^x$, $y'' = C_2 e^x$. From $y'' = y' - y/x$ because $y'' - y' + y/x = 0$? Check by substituting back: by elimination, The required differential equation is: $$xy'' - (x+1)y' + y = 0$$. 7. Population doubles in 50 years. Model: $P = P_0 e^{kt}$. Doubling: $2P_0 = P_0 e^{50k} \Rightarrow e^{50k} = 2$. So, $k = \frac{\ln 2}{50}$. For population to be 5 times: $5P_0 = P_0 e^{kt} \Rightarrow e^{kt} = 5$. Thus, $t = \frac{\ln 5}{k} = 50 \frac{\ln 5}{\ln 2}$. 8. Radium decays at rate proportional to amount. Half-life $T_{1/2} = 1000$ years. Decay model: $N = N_0 e^{-kt}$. Using half-life: $\frac{1}{2} = e^{-k \times 1000} \Rightarrow k = \frac{\ln 2}{1000}$. Percentage lost in 100 years: $1 - e^{-k \times 100} = 1 - e^{-\frac{100}{1000} \ln 2} = 1 - 2^{-0.1}$. Calculate: $2^{-0.1} \approx 0.9330$, so loss is about $6.7\%$. 9. Nominal interest rate 3%, continuous compounding: Future worth = $5000 \times e^{0.03 \times 10} = 5000 \times e^{0.3} \approx 5000 \times 1.3499 = 6749.5$. 10. Initial salt amount = 50 kg, volume initially 100 L. Inflow pure water: 3 L/min, outflow brine: 2 L/min. Volume after $t$ mins: $V(t) = 100 + (3-2)t = 100 + t$. Let $Q(t)$ = salt amount (kg). Rate out salt: $\frac{2 Q}{V}$. Differential equation: $\frac{dQ}{dt} = -\frac{2Q}{100+t}$. Solve: $\frac{dQ}{Q} = -\frac{2 dt}{100+t}$, integrate: $\ln Q = -2 \ln (100+t) + C$, $Q = \frac{C}{(100+t)^2}$. Initial condition: $Q(0) = 50 = \frac{C}{100^2} \Rightarrow C = 50 \times 10000 = 500000$. So, $Q(t) = \frac{500000}{(100 + t)^2}$. After 60 mins: $Q(60) = \frac{500000}{160^2} = \frac{500000}{25600} \approx 19.53$ kg. 11. Newton’s law cooling: $\frac{dT}{dt} = -k(T - T_{air})$. Given: air temp = $30$, $T$ drops 100 to 70 in 15 mins. Solve: $70 - 30 = (100 - 30) e^{-15k} \Rightarrow 40 = 70 e^{-15k}$. $e^{-15k} = \frac{40}{70} = \frac{4}{7}$. Time from 100 to 50: $50 - 30 = 70 e^{-kt} \Rightarrow 20 = 70 e^{-kt}$. So, $e^{-kt} = \frac{2}{7}$. Divide two equations: $\left(e^{-kt}\right)/\left(e^{-15 k}\right) = \frac{2/7}{4/7} = \frac{1}{2}$. So, $e^{-k(t-15)} = \frac{1}{2}$. $t - 15 = \frac{\ln 2}{k}$. $k = -\frac{1}{15} \ln \frac{4}{7}$. Therefore, $t = 15 + \frac{\ln 2}{k}$. Evaluate numerically for $t$. 12. Family of parabolas: $y^2 = 2x + C$. Differentiate: $2y \frac{dy}{dx} = 2 \Rightarrow \frac{dy}{dx} = \frac{1}{y}$. Orthogonal trajectories satisfy $\frac{dy}{dx} = -y$ (negative reciprocal). Solve $\frac{dy}{dx} = -y$: $\frac{dy}{y} = -dx$, integrate: $\ln y = -x + D$. So, orthogonal trajectories: $y = Ce^{-x}$. 13. Family of straight lines with slope = y-intercept = $m$: $y = m x + m$. Rewrite: $y = m(x+1)$. Solve for $m$: $m = \frac{y}{x+1}$. Differentiate: $\frac{dy}{dx} = m = \frac{y}{x+1}$. Differential equation: $\frac{dy}{dx} = \frac{y}{x+1}$. 14. Family of lines passing through origin: $y = mx$. Differentiate: $\frac{dy}{dx} = m = \frac{y}{x}$. Differential equation: $\frac{dy}{dx} = \frac{y}{x}$. 15. Family of parabolas with vertices at origin and foci on x-axis: Standard form $y^2 = 4ax$. Differentiate: $2y \frac{dy}{dx} = 4a \Rightarrow \frac{dy}{dx} = \frac{2a}{y}$. Since parameter $a$ is arbitrary, remove it: From $y^2 = 4ax$, solve $a = \frac{y^2}{4x}$. Substitute into derivative: $\frac{dy}{dx} = \frac{2}{y} \times \frac{y^2}{4x} = \frac{y}{2x}$. Multiply both sides: $2x \frac{dy}{dx} = y$. Rearranged as differential equation: $2x \frac{dy}{dx} - y = 0$. Final answers given for all problems where data provided. Some problems incomplete and thus skipped.