1.Answer: Square both sides: $(y'')^2 = (1 + (y')^2)^3$. Order = 2, Degree = 2.
2.Answer: Rearrange: $\frac{d^2y}{dx^2} = \sin(x+y)$. Order = 2, Degree = 1.
3.Answer: $y = C_1 \cos(x+c_3) + C_2 e^x$. The constants $c_1, c_2$ combine, and $c_4, c_5$ combine. Total independent constants: 3. Order = 3.
4.Answer: Eq: $(x-a)^2 + (y-a)^2 = a^2 \implies x^2 + y^2 - 2ax - 2ay + a^2 = 0$. Differentiate to eliminate $a$: $(x-y)^2(1+(y')^2) = (x+yy')^2$.
5.Answer: $2xy dy - (x^2+y^2) dx = dx \implies \frac{2xy dy - y^2 dx}{x^2} = \frac{x^2+1}{x^2} dx \implies d(y^2/x) = (1 + 1/x^2) dx \implies y^2/x = x - 1/x + C$.
6.Answer: Let $x+y = v \implies 1 + y' = v'$. $v' - 1 = \sin v + \cos v \implies \int \frac{dv}{1 + \sin v + \cos v} = x + C$. Use $\tan(v/2)$ substitution. Result: $\ln|1 + \tan(\frac{x+y}{2})| = x + C$.
7.Answer: Sub $y=vx \implies v + x v' = v + \sqrt{1+v^2} \implies \int \frac{dv}{\sqrt{1+v^2}} = \int \frac{dx}{x} \implies \ln|v + \sqrt{1+v^2}| = \ln|x| + \ln C \implies y + \sqrt{x^2+y^2} = Cx^2$.
8.Answer: Homogeneous. Sub $y=vx \implies x v' = \frac{-1-v^2-3v^2}{3v} = \frac{-1-4v^2}{3v} \implies \int \frac{3v}{1+4v^2} dv = -\int \frac{dx}{x} \implies \frac{3}{8}\ln(1+4v^2) = -\ln|x| + C \implies x^8(1+4y^2/x^2)^3 = K$.
9.Answer: $\frac{dx}{dy} - \frac{1}{y}x = 2y^2$. Linear in $x$. I.F. = $e^{\int -1/y dy} = 1/y$.
10.Answer: Divide by $y^3 \implies y^{-3}y' + y^{-2}\tan x = \sec x$. Sub $v=y^{-2}$. Linear DE yields $y^{-2} \sec^2 x = 2\sec x + C$.
11.Answer: $y' = -y/x$. Orthogonal $y' = x/y \implies y dy = x dx \implies y^2 - x^2 = C$. (Hyperbolas).
12.Answer: Homogeneous. Result: $y + \sqrt{x^2+y^2} = Cx^2$.
13.Answer: $e^{-4y} dy = e^{3x} dx \implies -\frac{1}{4}e^{-4y} = \frac{1}{3}e^{3x} + C$. Use $(0,0) \implies C = -7/12$. Solution: $4e^{3x} + 3e^{-4y} = 7$.
14.Answer: $\frac{dx}{dy} - x \cos y = \sin 2y$. Linear in $x$. I.F. = $e^{-\sin y}$. $x e^{-\sin y} = \int 2\sin y \cos y e^{-\sin y} dy = -2e^{-\sin y}(\sin y + 1) + C$.
15.Answer: $y = y_1 + C(y_1 - y_2)$ where $C$ is a constant.
16.Answer: Normal eq: $Y-y = -\frac{1}{y'}(X-x)$. $G$ is $(x+yy', 0)$. $OG^2 = x^2+y^2 \implies (x+yy')^2 = x^2+y^2 \implies x+yy' = \pm \sqrt{x^2+y^2}$.
17.Answer: Sub $v = \log y \implies v' = \frac{1}{y}y'$. Equation becomes $v' + \frac{1}{x}v = \frac{1}{x^2}v^2$. Bernoulli in $v$. Solution: $\frac{1}{\log y} = \frac{1}{2x} + Cx$.
18.Answer: Char eq: $m^2 - 5m + 6 = 0 \implies (m-2)(m-3)=0$. $m=2, 3$.
19.Answer: Homogeneous. $x v' = \frac{1+\cos v}{x}$. This requires more careful grouping. Result: $\sec(y/x) + \tan(y/x) = -1/x + C$.
20.Answer: $x^2 dx + y^2 dx + 2x dx + 2y dy = 0$. Group: $d(x^3/3 + x y^2 + x^2) = 0$? No. $(x+1)^2 + y^2 = C e^{-x}$? Direct integration after dividing by proper factor. Result: $e^x(x^2+y^2)=C$.
21.Answer: $2xy^2 dx + (y e^x dx - e^x dy) = 0 \implies 2x dx + \frac{y e^x dx - e^x dy}{y^2} = 0 \implies 2x dx + d(e^x/y) = 0 \implies x^2 + e^x/y = C$.
22.Answer: Family: $x^2 + (y-b)^2 = b^2 \implies x^2+y^2-2by=0 \implies y' = \frac{2xy}{x^2-y^2}$. Orthogonal: $y' = \frac{y^2-x^2}{2xy} \implies (x-c)^2 + y^2 = c^2$. (Circles touching y-axis at origin).
23.Answer: Homogeneous. $x^2+y^2=Cx$.
24.Answer: Area $\frac{1}{2}y \cdot \text{subtangent} = k \implies \frac{1}{2}y \cdot |y/y'| = k \implies y^2 = 2k y'$. Integrating gives $y = \frac{2k}{C-x}$ or similar hyperbolas.
25.Answer: Bernoulli. Sub $v = \sqrt{y} \implies 2v' + \frac{x}{1-x^2}v = x$. Result: $\sqrt{y} = (1-x^2)^{1/4} [ \int x(1-x^2)^{-1/4} dx + C]$.
26.Answer: $(x^2-y^2) dy + 2xy dx = 0$.
27.Answer: I.F. = $1+x^2$. $y(1+x^2) = \int 4x^2 dx = \frac{4x^3}{3} + C$.
28.Answer: Homogeneous. $x/y + \log|y| = C$.
29.Answer: Sub $x+y=v$. Becomes $v' - 1 = \frac{v+1}{2v+3} \implies \frac{2v+3}{3v+4} dv = dx$. Solve and sub back.
30.Answer: Exact check: $\frac{\partial M}{\partial y} = 2x+1-\sec^2 y$. $\frac{\partial N}{\partial x} = 2x-\tan^2 y$. Note $1+\tan^2 y = \sec^2 y$. They match. Solution: $x^2y + xy - x \tan y + \tan y = C$.
31.Answer: Homogeneous. $\tan(y/x) = \log|x| + 1$.
32.Answer: Linear in $x$. $\frac{dx}{dy} - \frac{1}{y}x = y^{-2/3}$. I.F. = $1/y$. $x/y = \int y^{-5/3} dy = -3/2 y^{-2/3} + C$.
33.Answer: Bernoulli. Sub $v = y^{-3}$. Result: $y^{-3} x^3 = 3\sin x + C$.
34.Answer: Logistic growth: $\frac{dP}{dt} = kP(L-P)$.
35.Answer: Sub $v = \log y$. $v' + v/x = v^2/x^2$. Reducible to linear. Result: $\frac{1}{\log y} = \frac{1}{2x} + Cx$.
36.Answer: Exact DE. Becomes $d(x y \cos y + x^2/2 \sin y) \dots$? No. Result: $x \sin y + y \cos y = C$.
37.Answer: I.F. = $1/x^2$. $y/x^2 = \int x dx = x^2/2 + C \implies y = x^4/2 + Cx^2$.
38.Answer: $y \frac{dy}{dx} = k \implies y^2 = 2kx + C$. (Parabolas).
39.Answer: $y dx + x dy + x y^2 dx - x^2 y dy = 0 \implies d(xy) + xy(y dx - x dy) = 0 \implies \frac{d(xy)}{(xy)^2} + \frac{y dx - x dy}{xy} = 0$. Result: $1/(xy) + \log(x/y) = C$.
40.Answer: I.F. = $e^{\tan^{-1}x}$. $y e^{\tan^{-1}x} = \int \frac{e^{2\tan^{-1}x}}{1+x^2} dx = \frac{1}{2} e^{2\tan^{-1}x} + C$.
41.Answer: $e^y = e^x + x^3/3 + C$. $(0,0) \implies 1 = 1 + 0 + C \implies C=0$. $y = \ln(e^x + x^3/3)$.
42.Answer: Linear in $\cos y$ or similar. Sub $v = \cos y$? No. Rewrite: $\frac{dy}{dx} + y \cot x = \dots$ actually divide by $\sin y$. Result: $\cos y = \sin x + C \csc x$.
43.Answer: Homogeneous. $(x^2-y^2)^2 = Cx y$.
44.Answer: Diff eq: $y' = 2a/y$. Eliminate $a$: $y (y')^2 + 2x y' - y = 0$. Orthogonal: Replace $y'$ with $-1/y'$. Same DE. Hence self-orthogonal.
45.Answer: I.F. = $e^x$. $y e^x = \int \frac{1+x}{x} dx = \ln|x| + x + C \implies y = e^{-x}(\ln|x| + x + C)$.
46.Answer: Homogeneous. $-e^{-y/x} = \ln|x| + C$.
47.Answer: Eq: $y^2 = 4a(x-h)$. Two constants. $y (y'') + (y')^2 = 0$.
48.Answer: Sub $v = \tan y$. $v' + 2xv = x^3$. I.F. = $e^{x^2}$. $v e^{x^2} = \int x^3 e^{x^2} dx$. Result: $\tan y = \frac{1}{2}(x^2-1) + C e^{-x^2}$.
49.Answer: $y' + \frac{1}{1+x^2}y = \frac{e^x}{1+x^2}$. I.F. = $e^{\tan^{-1}x}$. $y e^{\tan^{-1}x} = \int \frac{e^{x+\tan^{-1}x}}{1+x^2} dx$. $(0,0) \implies C = -1$.
50.Answer: Bernoulli. $1/y = \log x + 1 + Cx$.
51.Answer: Homogeneous. $\sin(y/x) = Cx$.
52.Answer: $\frac{dy}{y-1} = \frac{dx}{x(x+1)} \implies \ln|y-1| = \ln|x| - \ln|x+1| + C \implies y-1 = \frac{Cx}{x+1}$. $(1,0) \implies -1 = C/2 \implies C=-2$. Curve: $y = 1 - \frac{2x}{x+1} = \frac{1-x}{1+x}$.
53.Answer: Homogeneous. $e^{x^2/2y^2} = Cx$.
54.Answer: I.F. = $e^{\tan x}$. $y = \tan x - 1 + C e^{-\tan x}$.
55.Answer: I.F. = $\sqrt{1+x^2}$. $y\sqrt{1+x^2} = \tan^{-1}x + C$.
56.Answer: Exact. $x e^y + y^2 = C$.
57.Answer: $r = c(1 - \cos \theta)$.
58.Answer: $\phi(y/x) = Cx$.
59.Answer: I.F. = $x$. $xy = \frac{x^{n+2}}{n+2} + C$.
60.Answer: $\log(x-y) = x+y+1$.
