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Answer: Rearrange: $(y - x \frac{dy}{dx})^2 = 1 + (\frac{dy}{dx})^2$. Order = 1, Degree = 2.

Answer: Order = 3. Degree is Not Defined (since $\frac{d^3y}{dx^3}$ is in an exponential).

Answer: Equation: $x^2 + (y-b)^2 = b^2 \implies x^2 + y^2 - 2by = 0$. One arbitrary constant ($b$). Order = 1.

Answer: Square both sides: $[1 + (\frac{dy}{dx})^2]^3 = \rho^2 (\frac{d^2y}{dx^2})^2$. Order = 2, Degree = 2.

Answer: Equation $y^2 = 4ax$. Diff: $2y \frac{dy}{dx} = 4a$. Sub $4a$ into original: $y^2 = (2y \frac{dy}{dx})x \implies y = 2x \frac{dy}{dx}$.

Answer: $y' = -a\sin(x+b)$, $y'' = -a\cos(x+b) = -y \implies y'' + y = 0$.

Proof: Differentiating twice and substituting into the equation yields zero. Standard homogeneous second-order result.

Answer: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Differentiating twice and eliminating $a, b$: $xy \frac{d^2y}{dx^2} + x(\frac{dy}{dx})^2 - y \frac{dy}{dx} = 0$.

Answer: $\frac{d^2y}{dx^2} = e^x$. Order = 2, Degree = 1.

Answer: $y = (c_1+c_2+c_3 e^{c_4}) e^x$. This is just $y = A e^x$. One constant ($A$). Order = 1.

Answer: $\int \frac{dy}{1+y^2} = \int \frac{dx}{1+x^2} \implies \tan^{-1}y = \tan^{-1}x + C$.

Answer: $y = \int \sin^{-1} x dx = x\sin^{-1}x + \sqrt{1-x^2} + C$.

Answer: $\frac{dy}{1+y^2} = -\frac{e^x}{1+(e^x)^2} dx \implies \tan^{-1}y = -\tan^{-1}(e^x) + C$.

Answer: $\frac{dy}{y} = \tan x dx \implies \ln|y| = \ln|\sec x| + C$. Using $(0,1) \implies \ln 1 = \ln 1 + C \implies C=0$. $y = \sec x$.

Answer: Let $4x+y+1=v \implies 4+y'=v'$. $v'-4 = v^2 \implies \int \frac{dv}{v^2+4} = \int dx \implies \frac{1}{2}\tan^{-1}(\frac{4x+y+1}{2}) = x + C$.

Answer: $\int \frac{dy}{\sqrt{1-y^2}} = -\int \frac{dx}{\sqrt{1-x^2}} \implies \sin^{-1}y = -\sin^{-1}x + C \implies \sin^{-1}y + \sin^{-1}x = C$.

Answer: $dy = (2x + 1/x) dx \implies y = x^2 + \ln|x| + C$. Using $(1,1) \implies 1 = 1 + 0 + C \implies C=0$. Curve: $y = x^2 + \ln|x|$.

Answer: $\frac{dy}{dx} = e^{-y}(e^x + x^2) \implies e^y dy = (e^x + x^2) dx \implies e^y = e^x + \frac{x^3}{3} + C$.

Answer: $\cot y dy = -\tan x dx \implies \ln|\sin y| = \ln|\cos x| + \ln C \implies \sin y = C \cos x$.

Answer: $(\sin y + y\cos y) dy = (2x\log x + x) dx$. Integrate both: $y\sin y = x^2 \log x + C$.

Answer: Let $x+y=v \implies 1+y'=v'$. $v' - 1 = \cos v + \sin v$. $\int \frac{dv}{1+\cos v + \sin v} = x+C$. Final involves $\tan(v/2)$.

Answer: $\frac{y^2}{y+1} dy = -\frac{x^2}{x-1} dx \implies (y-1+\frac{1}{y+1}) dy = -(x+1+\frac{1}{x-1}) dx$. Integrate: $\frac{y^2}{2}-y+\ln|y+1| = -(\frac{x^2}{2}+x+\ln|x-1|) + C$.

Answer: $y=vx \implies v + x \frac{dv}{dx} = \frac{v^2-1}{2v} \implies x \frac{dv}{dx} = \frac{-1-v^2}{2v} \implies \int \frac{2v}{1+v^2} dv = -\int \frac{dx}{x} \implies x(1+v^2)=C \implies x^2+y^2=Cx$.

Answer: $y=vx \implies v + x \frac{dv}{dx} = 1 - 2v^2 + v \implies x \frac{dv}{dx} = 1 - 2v^2 \implies \int \frac{dv}{1-2v^2} = \ln|x| + C \implies \frac{1}{2\sqrt{2}}\ln|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}| = \ln|x| + C$.

Answer: $y=vx \implies x \frac{dv}{dx} = \tan v \implies \cot v dv = \frac{dx}{x} \implies \ln|\sin v| = \ln|x| + C \implies \sin(y/x) = Cx$.

Answer: $\cos v (v+x v') = v\cos v + 1 \implies x \cos v v' = 1 \implies \int \cos v dv = \int \frac{dx}{x} \implies \sin(y/x) = \ln|x| + C$.

Answer: $x^2y dx = (x^3+y^3) dy \implies \frac{dx}{dy} = \frac{x^3+y^3}{x^2y} = \frac{x}{y} + \frac{y^2}{x^2}$. Let $x=vy \implies v + y v' = v + 1/v^2 \implies v^2 dv = \frac{dy}{y} \implies \frac{v^3}{3} = \ln|y| + C \implies x^3 = 3y^3 \ln|y| + Cy^3$.

Answer: $\frac{dx}{dy} = \frac{x(2-\log(y/x))}{y} = \frac{x}{y}(2+\log(x/y))$. Let $x=vy \implies v + y v' = v(2+\log v) \implies \frac{dv}{v(1+\log v)} = \frac{dy}{y} \implies \ln|1+\log(x/y)| = \ln|y| + C$.

Answer: $y=vx \implies x v' = \frac{\sin v}{\cos v} \implies \cot v dv = \frac{dx}{x} \implies \ln|\sin v| = \ln|x| + C \implies \sin(y/x) = Cx$.

Answer: $\frac{dy}{dx} = \frac{y^2 - x\sqrt{x^2+y^2}}{xy} = \frac{y}{x} - \frac{\sqrt{x^2+y^2}}{y} = v - \frac{\sqrt{1+v^2}}{v}$. Separating leads to $\int \frac{v}{\sqrt{1+v^2}} dv = -\int \frac{dx}{x} \implies \sqrt{1+(y/x)^2} = -\ln|x| + C$.

Answer: $x v' = -\sin v \implies \csc v dv = -\frac{dx}{x} \implies \ln|\csc v - \cot v| = -\ln|x| + C \implies x(\csc(y/x) - \cot(y/x)) = K$.

Answer: $x v' = \frac{1+v^2}{1+v} - v = \frac{1-v}{1+v} \implies \int \frac{1+v}{1-v} dv = \ln|x| + C \implies -v - 2\ln|1-v| = \ln|x| + C$. Using $(1,1)$ where $y=1, x=1 \implies v=1$... wait, division by zero. Correct approach: use $y=1, x=1 \implies v=1$. Actually integral is $\ln|x(1-v)^2| + v = C$. Result: $\ln|x(1-y/x)^2| + y/x = 1$.

Answer: Let $x=vy \implies \frac{dx}{dy} = v + y \frac{dv}{dy}$. Becomes $2 v' e^v y + 1 = 0 \implies 2e^v dv = -dy/y \implies 2e^{x/y} = -\ln|y| + C$.

Answer: $y=vx \implies x v' = 1+v \implies \frac{dv}{1+v} = \frac{dx}{x} \implies \ln|1+v| = \ln|x| + C \implies 1+y/x = Cx \implies x+y = Cx^2$.

Answer: $x v' = \frac{v}{1+v^2} - v = \frac{-v^3}{1+v^2} \implies \int \frac{1+v^2}{v^3} dv = -\ln|x| + C \implies -1/(2v^2) + \ln|v| = -\ln|x| + C \implies \ln|y| - \frac{x^2}{2y^2} = C$.

Answer: $I.F. = e^{\int \sec x dx} = \sec x + \tan x$. $y(\sec x + \tan x) = \int \tan x (\sec x + \tan x) dx = \sec x + \tan x - x + C$.

Answer: $y' + \frac{2}{x}y = x \log x$. $I.F. = x^2$. $y x^2 = \int x^3 \log x dx = \frac{x^4}{4}\log x - \frac{x^4}{16} + C$.

Answer: $y' - \frac{x}{1-x^2}y = \frac{1}{1-x^2}$. $I.F. = e^{\int \frac{-x}{1-x^2} dx} = e^{\frac{1}{2}\ln|1-x^2|} = \sqrt{1-x^2}$.

Answer: $I.F. = 1+x^2$. $y(1+x^2) = \int \frac{4x^2}{1+x^2}(1+x^2) dx = \frac{4x^3}{3} + C$.

Answer: $I.F. = e^{-3\ln|\sin x|} = \csc^3 x$. $y \csc^3 x = \int 2\sin x \cos x \csc^3 x dx = \int 2\cot x \csc x dx = -2\csc x + C$. $y = -2\sin^2 x + C \sin^3 x$. $y(\pi/2)=2 \implies 2 = -2 + C \implies C=4$. Solution: $y = 4\sin^3 x - 2\sin^2 x$.

Answer: $\frac{dx}{dy} - x = y+1$. $I.F. = e^{-y}$. $x e^{-y} = \int (y+1)e^{-y} dy = -(y+1)e^{-y} - e^{-y} + C \implies x = -(y+2) + C e^y$.

Answer: $\frac{dx}{dy} - \frac{1}{y}x = 2y$. $I.F. = 1/y$. $x/y = \int 2 dy = 2y + C \implies x = 2y^2 + Cy$.

Answer: $I.F. = e^x$. $y e^x = \int e^x(\cos x - \sin x) dx = e^x \cos x + C \implies y = \cos x + C e^{-x}$.

Answer: $I.F. = e^{\int \frac{1}{x \log x} dx} = \log x$. $y \log x = \int \frac{2 \log^2 x}{x} dx = \frac{2}{3}(\log x)^3 + C$.

Answer: $I.F. = 1+x^2$. $y(1+x^2) = \int \cot x dx = \ln|\sin x| + C$.

Answer: $\tan^{-1}y = -\ln(1+e^x) + C$. Using $(0,1) \implies \pi/4 = -\ln 2 + C$. Result: $\tan^{-1}y = \ln(\frac{2e^{\pi/4}}{1+e^x})$.

Answer: $I.F. = y$. $xy = \int y^3 dy = y^4/4 + C$.

Answer: $\frac{dx}{dy} - \frac{1}{y}x = 2y^2$. $I.F. = 1/y$. $x/y = \int 2y dy = y^2 + C \implies x = y^3 + Cy$.

Answer: $y' + \sec^2 x y = \tan x \sec^2 x$. $I.F. = e^{\tan x}$. $y e^{\tan x} = \int e^{\tan x} \tan x \sec^2 x dx = e^{\tan x}(\tan x - 1) + C$.

Answer: $y' - \frac{1}{x}y = (1-1/x)e^x$. $I.F. = 1/x$. $y/x = \int (1/x - 1/x^2)e^x dx = \frac{e^x}{x} + C \implies y = e^x + Cx$.

Answer: $y^{-1}/x = \int \frac{\log x}{x^2} dx = -\frac{1}{x}\log x - \frac{1}{x} + C \implies 1/y = -\log x - 1 + Cx$.

Answer: Original: $y = ax^2 \implies dy/dx = 2y/x$. Orthogonal: $dy/dx = -x/2y \implies \int 2y dy = -\int x dx \implies y^2 = -x^2/2 + C \implies x^2 + 2y^2 = K$. (Ellipses).

Answer: $N = N_0 e^{kt}$. $2 = e^{5k} \implies 5k = \ln 2$. For triple: $3 = e^{kt} \implies kt = \ln 3 \implies t = 5 \frac{\ln 3}{\ln 2} \approx 7.9$ hours.

Answer: $\frac{T-20}{100-20} = e^{-10k}$. $\frac{40}{80} = 1/2 = e^{-10k}$. For 20 mins: $\frac{T-20}{80} = e^{-20k} = (1/2)^2 = 1/4 \implies T-20 = 20 \implies T = 40^\circ C$.

Answer: $y \frac{dy}{dx} = x \implies \int y dy = \int x dx \implies y^2/2 = x^2/2 + C$. Using $(0,-2) \implies 4/2 = 0 + C \implies C=2$. Curve: $y^2 - x^2 = 4$.

Answer: $1/y = \log x + 1 + Cx$.

Answer: Original: $x^2+y^2=c^2 \implies y' = -x/y$. Orthogonal: $y' = y/x \implies y = kx$ (Family of lines through origin).

Answer: $dP/dt = 0.05 P \implies P = P_0 e^{0.05t}$. $2 = e^{0.05t} \implies 0.05t = \ln 2 \implies t = 20 \ln 2 \approx 13.86$ years.

Answer: $1/y \cos x = x + C$.

Answer: $dy/dx = 2y/x \implies \frac{dy}{y} = \frac{2dx}{x} \implies y = Cx^2$. Through $(1,1) \implies 1 = C(1) \implies C=1$. Curve: $y = x^2$.