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Answer: Order is 2 (highest derivative is $\frac{d^2y}{dx^2}$). Degree is 3 (power of the highest derivative).

Answer: Degree is Not Defined. (The differential equation cannot be written as a polynomial equation in its derivatives due to $\sin(\frac{dy}{dx})$).

Answer: Order is 2. (There are 2 arbitrary constants, $A$ and $B$).

Proof: If $y = e^x$, then $y' = e^x$ and $y'' = e^x$. Substitute into $y'' - y = 0 \implies e^x - e^x = 0$. Yes, it is a solution.

Answer: $y = mx \implies \frac{dy}{dx} = m$. Substitute $m$ back into the original equation: $y = \left(\frac{dy}{dx}\right)x \implies x \frac{dy}{dx} - y = 0$.

Answer: 3 (The number of arbitrary constants equals the order of the general solution).

Answer: 0 (A particular solution contains no arbitrary constants).

Answer: Order is 1 (highest derivative is $\frac{dy}{dx}$). Degree is 2.

Answer: Square both sides: $1 + \left(\frac{dy}{dx}\right)^3 = \left(\frac{d^2y}{dx^2}\right)^2$. Order is 2, and its Degree is 2.

Proof: $y = \sin x + \cos x \implies y' = \cos x - \sin x \implies y'' = -\sin x - \cos x$. Substitute into $y'' + y = 0 \implies (-\sin x - \cos x) + (\sin x + \cos x) = 0$. Yes, it is a solution.

Answer: $y \,dy = x \,dx \implies \int y \,dy = \int x \,dx \implies \frac{y^2}{2} = \frac{x^2}{2} + C \implies y^2 - x^2 = 2C$ (or $C'$).

Answer: $\frac{dy}{dx} = e^x \cdot e^y \implies e^{-y} \,dy = e^x \,dx \implies \int e^{-y} \,dy = \int e^x \,dx \implies -e^{-y} = e^x + C \implies e^x + e^{-y} = K$.

Answer: $\frac{dy}{1+y^2} = (1+x^2) \,dx \implies \int \frac{dy}{1+y^2} = \int (1+x^2) \,dx \implies \tan^{-1} y = x + \frac{x^3}{3} + C$.

Answer: $\frac{dy}{y} = \frac{dx}{x} \implies \ln|y| = \ln|x| + \ln C \implies y = Cx$.

Answer: $dy = \sin x \,dx \implies \int dy = \int \sin x \,dx \implies y = -\cos x + C$.

Answer: Substitute $v = x+y$ (which gives $\frac{dv}{dx} = 1 + \frac{dy}{dx}$).

Answer: $\frac{dy}{y^2} = -4x \,dx \implies -\frac{1}{y} = -2x^2 + C$. Use $y(0)=1 \implies -1 = 0 + C \implies C = -1$. Solution: $-\frac{1}{y} = -2x^2 - 1 \implies y = \frac{1}{2x^2+1}$.

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

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

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

Answer: Yes. (Both numerator and denominator are homogeneous of degree 2, making the ratio homogeneous of degree 0).

Answer: $y = vx$ (or $v = \frac{y}{x}$).

Answer: $\frac{x+vx}{x} = 1+v$. Equating gives $v + x \frac{dv}{dx} = 1 + v \implies x \frac{dv}{dx} = 1$.

Answer: $\frac{dy}{dx} = \frac{x+y}{x}$. Sub $y=vx \implies x \frac{dv}{dx} = 1 \implies dv = \frac{dx}{x} \implies v = \ln|x| + C \implies \frac{y}{x} = \ln|x| + C \implies y = x\ln|x| + Cx$.

Answer: Yes. ($f(\lambda x, \lambda y) = \sin\left(\frac{\lambda y}{\lambda x}\right) = \sin\left(\frac{y}{x}\right) = \lambda^0 f(x,y)$).

Answer: Sub $y=vx \implies v + x \frac{dv}{dx} = v + \sin v \implies x \frac{dv}{dx} = \sin v \implies \csc v \,dv = \frac{dx}{x} \implies \ln|\csc v - \cot v| = \ln|x| + \ln C \implies \csc\left(\frac{y}{x}\right) - \cot\left(\frac{y}{x}\right) = Cx$.

Answer: $\frac{dy}{dx} = \frac{x^2+y^2}{2xy}$. Sub $y=vx \implies v + x \frac{dv}{dx} = \frac{1+v^2}{2v} \implies x \frac{dv}{dx} = \frac{1-v^2}{2v} \implies \frac{2v}{1-v^2} dv = \frac{dx}{x} \implies -\ln|1-v^2| = \ln|x| + C \implies x(1-v^2) = K \implies x^2 - y^2 = Cx$.

Answer: $x = vy$ (or $v = \frac{x}{y}$).

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

Answer: No. (The degree of the RHS is 1, not 0, so it cannot be solved directly with standard $y=vx$ substitution giving a separable form without extra $x$ terms).

Answer: $\frac{dy}{dx} + Py = Q$.

Answer: $P = 2$. I.F. = $e^{\int 2 \,dx} = e^{2x}$.

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

Answer: $P = \frac{1}{x}$. I.F. = $e^{\int \frac{1}{x} \,dx} = e^{\ln x} = x$.

Answer: Solution: $y(x) = \int x^2(x) \,dx = \int x^3 \,dx \implies xy = \frac{x^4}{4} + C$.

Answer: Divide by $x \implies \frac{dy}{dx} + \frac{2}{x}y = x^2$. $P = \frac{2}{x}$. I.F. = $e^{\int \frac{2}{x} \,dx} = e^{2\ln x} = e^{\ln x^2} = x^2$.

Answer: $P = \cot x$. I.F. = $e^{\int \cot x \,dx} = e^{\ln|\sin x|} = \sin x$.

Answer: Linear in $x$. $P_1 = \frac{1}{y}$. I.F. = $e^{\int \frac{1}{y} \,dy} = e^{\ln y} = y$.

Answer: Divide by $(1+x^2) \implies P = \frac{2x}{1+x^2}$. I.F. = $e^{\int \frac{2x}{1+x^2} \,dx} = e^{\ln(1+x^2)} = 1+x^2$.

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

Answer: $\frac{dy}{dx} + Py = Qy^n$ (where $P, Q$ are functions of $x$).

Answer: $y^{-2}\frac{dy}{dx} + \frac{1}{x}y^{-1} = 1$. Substitute $v = y^{-1}$ (which gives $\frac{dv}{dx} = -y^{-2}\frac{dy}{dx}$).

Answer: $\frac{dP}{dt} = kP$ (where $k$ is the proportionality constant).

Answer: $\frac{dP}{P} = 0.05 \,dt \implies \ln|P| = 0.05t + C \implies P(t) = P_0 e^{0.05t}$.

Answer: $\frac{dT}{dt} = -k(T - T_s)$ (where $k > 0$).

Answer: Let $v = \tan y \implies \frac{dv}{dx} = \sec^2 y \frac{dy}{dx}$.

Answer: Replace $\frac{dy}{dx}$ with $-\frac{dx}{dy}$. Orthogonal DE: $\frac{dy}{dx} = -\frac{1}{f(x,y)}$.

Answer: Orthogonal DE: $\frac{dy}{dx} = -\frac{1}{2x} \implies \int dy = \int -\frac{1}{2x} dx \implies y = -\frac{1}{2}\ln|x| + C$.

Answer: $\frac{dy}{dx} = \frac{y}{x} \implies \frac{dy}{y} = \frac{dx}{x} \implies \ln|y| = \ln|x| + C \implies y = Kx$ (Family of lines through origin).

Answer: Divide by $y^3$ (to obtain $y^{-3}\frac{dy}{dx} + y^{-2} = x$, then substitute $v = y^{-2}$).