1. Determine the order and degree of a differential equation (D.E.): The order is the highest derivative present in the equation. The degree is the power of the highest order derivative, provided the equation is polynomial in derivatives. 2. Given $\frac{dy}{dx} = x^2$, find $y$ passing through $(1,1)$. Integrate both sides: $$ y = \int x^2 \, dx = \frac{x^3}{3} + C $$ Using initial condition $y(1) = 1$: $$ 1 = \frac{1}{3} + C \implies C = \frac{2}{3} $$ So, $$ y = \frac{x^3}{3} + \frac{2}{3} $$ 3. Solve: (Problem text missing details; unable to solve) 4. Solve differential equation: $$ (\cos x \cos y - \cot x) dx - \sin x \sin y dy = 0 $$ Rewrite as $$ \frac{dy}{dx} = \frac{\cos x \cos y - \cot x}{\sin x \sin y} $$ Use substitution or simplification. Dividing numerator and denominator by $\sin x \sin y$: $$ \frac{dy}{dx} = \cot y \cot x - \frac{\cot x}{\sin x \sin y} $$ After simplification and separable steps, or rewrite and integrate. Step: Multiply entire equation to put in form: $$ (\cos x \cos y - \cot x) dx = \sin x \sin y dy $$ Convert $\cot x = \frac{\cos x}{\sin x}$: $$ (\cos x \cos y - \frac{\cos x}{\sin x}) dx = \sin x \sin y dy $$ Multiply both sides by $\sin x$: $$ (\sin x \cos x \cos y - \cos x) dx = \sin^2 x \sin y dy $$ Set $u = \sin x$, try substitution or other methods to solve the exact or separable equation. 5. Solve equation (Details missing; unable to solve) 6. Find differential equation whose general solution is $y = C_1 x + C_2 e^x$. Differentiate once: $$ y' = C_1 + C_2 e^x $$ Differentiate twice: $$ y'' = C_2 e^x $$ Subtract: $$ y'' - y' + y = 0 $$ This simplifies to the differential equation: $$ y'' - y' = 0 $$ 7. Population doubles in 50 years, find years to be 5 times assuming exponential growth: Population $P = P_0 e^{kt}$ Given: $$ 2 = e^{50k} \implies k = \frac{\ln 2}{50} $$ Find $t$ for: $$ 5 = e^{kt} \implies t = \frac{\ln 5}{k} = \frac{50 \ln 5}{\ln 2} $$ 8. Radium decomposes proportional to amount. Half disappears in 1000 years. Decay formula: $$ N = N_0 e^{-kt} $$ Given: $$ \frac{1}{2} = e^{-1000k} \implies k = \frac{\ln 2}{1000} $$ Find percentage lost in 100 years: $$ \text{Remaining} = e^{-k \times 100} = e^{-\frac{100\ln 2}{1000}} = 2^{-0.1} $$ Percentage lost = $$ (1 - 2^{-0.1}) \times 100 \approx 6.7\% $$ 9. Nominal interest rate 3%, amount 5000, continuous compounding for 10 years: $$ A = P e^{rt} = 5000 e^{0.03 \times 10} = 5000 e^{0.3} $$ Calculate: $$ 5000 \times e^{0.3} \approx 5000 \times 1.3499 = 6749.5 $$ 10. Salt tank problem: Initial volume = 100 L, initial salt = 50 kg. Inflow rate = 3 L/min (pure water), outflow = 2 L/min. Volume changes as: $$ V(t) = 100 + (3 - 2)t = 100 + t $$ Let $Q(t)$ = amount of salt in kg. Rate of salt change: $$ \frac{dQ}{dt} = (\text{in rate}) - (\text{out rate}) $$ In rate = 0 (pure water), out rate = $$ \frac{2 Q}{V(t)} = \frac{2Q}{100+t} $$ So, $$ \frac{dQ}{dt} = - \frac{2Q}{100+t} $$ Solve: $$ \frac{dQ}{Q} = - \frac{2 dt}{100+t} $$ Integrate: $$ \ln |Q| = -2 \ln |100+t| + C = \ln C -2 \ln(100+t) $$ $$ Q = \frac{C}{(100+t)^2} $$ At $t=0$, $Q=50$: $$ 50 = \frac{C}{100^2} \implies C = 50 \times 10000 = 500000 $$ Find $Q(60)$: $$ Q(60) = \frac{500000}{160^2} = \frac{500000}{25600} \approx 19.53 \text{ kg} $$ 11. Newton's law cooling: $$ \frac{dT}{dt} = -k (T - T_a) $$ Where $T_a=30$, initial: $$ T(0) = 100 $$ Given $T(15) = 70$. Solution: $$ T - 30 = (100 - 30) e^{-kt} = 70 e^{-kt} $$ At $t=15$: $$ 40 = 70 e^{-15k} \implies e^{-15k} = \frac{40}{70} = \frac{4}{7} $$ $$ k = - \frac{1}{15} \ln \frac{4}{7} $$ Find $t$ when $T=50$: $$ 20 = 70 e^{-kt} \implies e^{-kt} = \frac{2}{7} $$ $$ -kt = \ln \frac{2}{7} \implies t = - \frac{1}{k} \ln \frac{2}{7} $$ Substitute $k$: $$ t = \frac{15 \ln \frac{2}{7}}{\ln \frac{4}{7}} \approx 31.15 \text{ minutes} $$ 12. Orthogonal trajectories to family $y^2 = 2x + C$. Differentiate implicitly: $$ 2y \frac{dy}{dx} = 2 \implies \frac{dy}{dx} = \frac{1}{y} $$ The slope of original family is $\frac{1}{y}$. Orthogonal trajectories have slope $$ m_o = -\frac{1}{m} = -y $$ Form the DE for orthogonal trajectories: $$ \frac{dy}{dx} = -y $$ Solve: $$ \frac{dy}{y} = - dx $$ Integrate: $$ \ln |y| = -x + C $$ $$ y = Ce^{-x} $$ 13. Differential equation of family of lines with slope and intercept equal: Let slope = intercept = $m$. Equation: $$ y = mx + m $$ Rearranged: $$ y - mx - m = 0 $$ Replace $m$ with $y - mx$ from slope formula $m = \frac{dy}{dx}$: $$ y - x \frac{dy}{dx} - \frac{dy}{dx} = 0 $$ So differential equation: $$ y - x y' - y' = 0 $$ 14. Differential equation of family of lines passing through origin: Equation: $$ y = mx $$ Replace $m$ by $y' = \frac{dy}{dx}$: $$ \frac{y}{x} = y' $$ Or $$ y' = \frac{y}{x} $$ 15. Differential equation of family of parabolas with vertex at origin and foci on x-axis: Standard form: $$ y^2 = 4ax $$ Replace parameter $a$ by $C$: $$ y^2 = 4Cx $$ Differentiate: $$ 2y y' = 4C $$ Replace $C = \frac{y^2}{4x}$: $$ 2y y' = 4 \times \frac{y^2}{4x} = \frac{y^2}{x} $$ Simplify: $$ 2y y' = \frac{y^2}{x} \implies 2x y y' = y^2 \implies 2x y' = y $$ Rearranged: $$ 2x \frac{dy}{dx} = y $$ Final differential equation: $$ 2x \frac{dy}{dx} - y = 0 $$