Variable separable type differential equations Questions in English

(C) Given the substitution $\frac{dy}{dx}=z$,we differentiate with respect to $x$ to get $\frac{d^2y}{dx^2}=\frac{dz}{dx}$.
Substituting these into the given differential equation $\frac{d^2y}{dx^2}-\frac{dy}{dx}=0$,we obtain $\frac{dz}{dx}-z=0$.
This is a first-order separable differential equation: $\frac{dz}{dx}=z$.
Separating the variables,we have $\frac{dz}{z}=dx$.
Integrating both sides,we get $\int \frac{dz}{z} = \int dx$,which results in $\log_{e}|z|=x+C_1$.
Taking the exponential of both sides,we get $|z|=e^{x+C_1} = e^{C_1} \cdot e^{x}$.
Letting $A = \pm e^{C_1}$,we obtain the solution $z=Ae^{x}$.

(B) Given,the slope of the tangent is $\frac{dy}{dx} = x + xy$.
Rearranging the terms,we have $\frac{dy}{dx} = x(1+y)$.
Separating the variables,we get $\frac{dy}{1+y} = x dx$.
Integrating both sides,we get $\int \frac{dy}{1+y} = \int x dx$,which gives $\ln(y+1) = \frac{x^2}{2} + C$.
Exponentiating both sides,$y+1 = e^{\frac{x^2}{2} + C} = e^C \cdot e^{\frac{x^2}{2}}$.
Let $e^C = A$,so $y = A e^{\frac{x^2}{2}} - 1$.
Since the curve passes through $(0,1)$,we substitute $x=0$ and $y=1$: $1 = A e^0 - 1 \Rightarrow 1 = A - 1 \Rightarrow A = 2$.
Thus,the equation of the curve is $y = 2 e^{\frac{x^2}{2}} - 1$.

(B) Given differential equation is $\tan(y) dx + \sec^2(y) \tan(x) dy = 0$.
Rearranging the terms to separate variables,we get:
$\sec^2(y) \tan(x) dy = -\tan(y) dx$
$\frac{\sec^2(y)}{\tan(y)} dy = -\frac{1}{\tan(x)} dx$
$\frac{\sec^2(y)}{\tan(y)} dy = -\cot(x) dx$
Integrating both sides:
$\int \frac{\sec^2(y)}{\tan(y)} dy = -\int \cot(x) dx$
Let $u = \tan(y)$,then $du = \sec^2(y) dy$.
$\int \frac{1}{u} du = -\ln|\sin(x)| + C$
$\ln|\tan(y)| = -\ln|\sin(x)| + C$
$\ln|\tan(y)| + \ln|\sin(x)| = C$
$\ln|\sin(x) \tan(y)| = C$
Taking exponential on both sides,we get $\sin(x) \tan(y) = e^C = c$.

(A) Given differential equation is $\frac{dx}{dy} + 2yx = 2y$.
Rearranging the terms,we get $\frac{dx}{dy} = 2y(1-x)$.
Separating the variables,we have $\int \frac{dx}{1-x} = \int 2y \, dy$.
Integrating both sides,we get $-\ln|1-x| = y^2 + C$,which can be written as $\ln|1-x| = -y^2 - C$.
Taking the exponential of both sides,$|1-x| = e^{-y^2 - C} = Ae^{-y^2}$,where $A = e^{-C}$.
This implies $1-x = \pm Ae^{-y^2}$,or $x-1 = Ke^{-y^2}$ where $K = \mp A$.
Since the curve passes through the point $(2,0)$,substitute $x=2$ and $y=0$:
$2-1 = Ke^{-(0)^2} \Rightarrow 1 = K(1) \Rightarrow K=1$.
Thus,the solution is $(x-1) = e^{-y^2}$.

(B) Given differential equation: $e^x y dx + e^x dy + x dx = 0$
We can rewrite this as: $d(ye^x) + x dx = 0$
Integrating both sides: $\int d(ye^x) + \int x dx = \int 0 dx$
$ye^x + \frac{x^2}{2} = C_1$
Multiplying by $2$: $2ye^x + x^2 = 2C_1$
Let $2C_1 = C$,so the solution is $2ye^x + x^2 = C$.

(D) Given the differential equation $(1+y^2) dx - xy dy = 0$ with the initial condition $y(1)=0$.
Rearranging the terms,we get $(1+y^2) dx = xy dy$.
Separating the variables,we have $\frac{dx}{x} = \frac{y dy}{1+y^2}$.
Integrating both sides,$\int \frac{dx}{x} = \int \frac{y dy}{1+y^2}$.
Let $t = 1+y^2$,then $dt = 2y dy$,so $y dy = \frac{1}{2} dt$.
Thus,$\ln|x| = \frac{1}{2} \ln|1+y^2| + C_1 = \ln|\sqrt{1+y^2}| + C_1$.
This implies $x = c\sqrt{1+y^2}$.
Using the condition $y(1)=0$,we substitute $x=1$ and $y=0$: $1 = c\sqrt{1+0^2} \implies c=1$.
So,$x = \sqrt{1+y^2}$,which squares to $x^2 = 1+y^2$ or $x^2 - y^2 = 1$.
This is the equation of a rectangular hyperbola where $a^2=1$ and $b^2=1$.
The eccentricity $e$ of a hyperbola is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1}{1}} = \sqrt{2}$.

(A) Given the differential equation: $\frac{dy}{dx} = \frac{x-y+3}{2(x-y)+5}$.
Let $x-y = u$. Then $1 - \frac{dy}{dx} = \frac{du}{dx}$,which implies $\frac{dy}{dx} = 1 - \frac{du}{dx}$.
Substituting this into the equation: $1 - \frac{du}{dx} = \frac{u+3}{2u+5}$.
Rearranging the terms: $\frac{du}{dx} = 1 - \frac{u+3}{2u+5} = \frac{2u+5-u-3}{2u+5} = \frac{u+2}{2u+5}$.
Separating the variables: $\int \frac{2u+5}{u+2} du = \int dx$.
Performing the division: $\int (2 + \frac{1}{u+2}) du = x + C$.
Integrating: $2u + \log|u+2| = x + C$.
Substituting $u = x-y$ back: $2(x-y) + \log|x-y+2| + C = x$.
Comparing this with the given form $x = 2(x-y) + \log(t) + c$,we identify $t = x-y+2$.

(A) Given differential equation is $\log \left(\frac{dy}{dx}\right) = ax + by$.
Taking exponential on both sides,we get $\frac{dy}{dx} = e^{ax + by} = e^{ax} \cdot e^{by}$.
Separating the variables,we have $e^{-by} dy = e^{ax} dx$.
Integrating both sides,we get $\int e^{-by} dy = \int e^{ax} dx$.
This results in $-\frac{1}{b} e^{-by} = \frac{1}{a} e^{ax} + C_1$.
Multiplying by $ab$,we get $-a e^{-by} = b e^{ax} + ab C_1$.
Rearranging the terms,we get $b e^{ax} + a e^{-by} = c$,where $c = -ab C_1$.
Therefore,option $A$ is correct.

(C) Given differential equation is $\frac{dy}{dx} = \frac{2x(y-2) + 1(y-2)}{2y(x-2) + 1(x-2)} = \frac{(2x+1)(y-2)}{(2y+1)(x-2)}$.
Separating the variables,we get $\frac{2y+1}{y-2} dy = \frac{2x+1}{x-2} dx$.
Rewrite the fractions: $\frac{2(y-2)+5}{y-2} dy = \frac{2(x-2)+5}{x-2} dx$.
This simplifies to $(2 + \frac{5}{y-2}) dy = (2 + \frac{5}{x-2}) dx$.
Integrating both sides: $\int (2 + \frac{5}{y-2}) dy = \int (2 + \frac{5}{x-2}) dx$.
$2y + 5 \log |y-2| = 2x + 5 \log |x-2| + C$.
Rearranging terms: $2(y-x) + 5 \log |y-2| - 5 \log |x-2| = C$.
$2(y-x) + 5 \log \left| \frac{y-2}{x-2} \right| = C$.

(D) Given the differential equation $\frac{dy}{dx} = e^{-2x}$.
Integrating both sides with respect to $x$:
$\int dy = \int e^{-2x} dx$
$y = -\frac{1}{2}e^{-2x} + C$ $(i)$
Given the initial condition $y(\log 2) = \frac{1}{16}$:
$\frac{1}{16} = -\frac{1}{2}e^{-2(\log 2)} + C$
Since $e^{-2\log 2} = e^{\log(2^{-2})} = 2^{-2} = \frac{1}{4}$:
$\frac{1}{16} = -\frac{1}{2}(\frac{1}{4}) + C$
$\frac{1}{16} = -\frac{1}{8} + C$
$C = \frac{1}{16} + \frac{2}{16} = \frac{3}{16}$
Substituting $C$ back into equation $(i)$:
$y = -\frac{1}{2}e^{-2x} + \frac{3}{16}$
$y = \frac{-8e^{-2x} + 3}{16} = \frac{3 - 8e^{-2x}}{16}$

(D) Given the differential equation: $y^{\prime} = \frac{y}{2x}$
Separating the variables,we get: $\frac{dy}{y} = \frac{dx}{2x}$
Integrating both sides: $\int \frac{1}{y} dy = \frac{1}{2} \int \frac{1}{x} dx$
$\ln|y| = \frac{1}{2} \ln|x| + C$
$\ln|y| = \ln|x^{1/2}| + C$
Taking the exponential of both sides: $y = k \sqrt{x}$,where $k = e^C$
Squaring both sides: $y^2 = k^2 x$
Let $k^2 = c$,then $y^2 = cx$
This equation represents a family of parabolas opening along the $x$-axis.

(A) Given differential equation is $\frac{d y}{d x}+\frac{\sin (2 x+y)}{\cos x}+2=0$ ....$(i)$
Let $2 x+y=t$. Differentiating with respect to $x$,we get $2+\frac{d y}{d x}=\frac{d t}{d x}$,so $\frac{d y}{d x}=\frac{d t}{d x}-2$.
Substituting this into equation $(i)$:
$(\frac{d t}{d x}-2) + \frac{\sin t}{\cos x} + 2 = 0$
$\frac{d t}{d x} + \frac{\sin t}{\cos x} = 0$
$\frac{d t}{\sin t} = -\frac{d x}{\cos x}$
$\operatorname{cosec} t \, dt = -\sec x \, dx$
Integrating both sides:
$\int \operatorname{cosec} t \, dt = -\int \sec x \, dx$
$\ln|\operatorname{cosec} t - \cot t| = -\ln|\sec x + \tan x| + C_1$
$\ln|\operatorname{cosec} t - \cot t| + \ln|\sec x + \tan x| = C_1$
$\ln|(\sec x + \tan x)(\operatorname{cosec} t - \cot t)| = C_1$
$(\sec x + \tan x)(\operatorname{cosec} t - \cot t) = e^{C_1} = C$
Substituting $t = 2x+y$ back,we get:
$(\sec x + \tan x)[\operatorname{cosec}(2x+y) - \cot(2x+y)] = C$.

(C) Given the differential equation: $(3x-4y)(dx-3dy)+(6dx-4dy)=0$
Rearranging the terms: $(3x-4y+6)dx - (3(3x-4y)+4)dy = 0$
$\frac{dy}{dx} = \frac{3x-4y+6}{3(3x-4y)+4}$
Let $t = 3x-4y$,then $\frac{dt}{dx} = 3-4\frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{1}{4}(3-\frac{dt}{dx})$
Substituting into the equation: $\frac{1}{4}(3-\frac{dt}{dx}) = \frac{t+6}{3t+4}$
$3-\frac{dt}{dx} = \frac{4t+24}{3t+4} \Rightarrow \frac{dt}{dx} = 3 - \frac{4t+24}{3t+4} = \frac{9t+12-4t-24}{3t+4} = \frac{5t-12}{3t+4}$
Separating variables: $dx = \frac{3t+4}{5t-12}dt$
$dx = (\frac{3}{5} + \frac{56/5}{5t-12})dt$
Integrating both sides: $x = \frac{3}{5}t + \frac{56}{25}\log|5t-12| + C$
Substitute $t = 3x-4y$: $x = \frac{3}{5}(3x-4y) + \frac{56}{25}\log|5(3x-4y)-12| + C$
$x = \frac{9x-12y}{5} + \frac{56}{25}\log|15x-20y-12| + C$
$25x = 45x-60y + 56\log|15x-20y-12| + 25C$
$60y-20x = 56\log|15x-20y-12| + C'$
Dividing by $4$: $15y-5x = 14\log|15x-20y-12| + C''$
$5x-15y+14\log|15x-20y-12|=C$.

(D) Given the differential equation: $\frac{dy}{dx} = \frac{xy+x-2y-2}{xy-2x+y-2}$
Factorizing the numerator and denominator:
$\frac{dy}{dx} = \frac{x(y+1)-2(y+1)}{x(y-2)+1(y-2)} = \frac{(x-2)(y+1)}{(x+1)(y-2)}$
Separating the variables:
$\frac{y-2}{y+1} dy = \frac{x-2}{x+1} dx$
Rewrite the fractions:
$\frac{y+1-3}{y+1} dy = \frac{x+1-3}{x+1} dx$
$(1 - \frac{3}{y+1}) dy = (1 - \frac{3}{x+1}) dx$
Integrating both sides:
$\int (1 - \frac{3}{y+1}) dy = \int (1 - \frac{3}{x+1}) dx$
$y - 3 \ln |y+1| = x - 3 \ln |x+1| + c$
Rearranging the terms:
$x - y + 3 \ln |y+1| - 3 \ln |x+1| = c$
$x - y + 3 \ln \left| \frac{y+1}{x+1} \right| = c$

(C) Given differential equation: $\frac{y^2 e^{-1/y}}{\sqrt{x}} dx = 2 \sec \sqrt{x} dy$.
Separate the variables: $\frac{dx}{\sqrt{x} \sec \sqrt{x}} = \frac{2 dy}{y^2 e^{-1/y}}$.
This simplifies to: $\cos \sqrt{x} \frac{dx}{\sqrt{x}} = 2 e^{1/y} \frac{dy}{y^2}$.
Integrate both sides: $\int \cos \sqrt{x} \frac{dx}{\sqrt{x}} = \int 2 e^{1/y} \frac{dy}{y^2}$.
Let $u = \sqrt{x}$,then $du = \frac{1}{2\sqrt{x}} dx \implies \frac{dx}{\sqrt{x}} = 2 du$.
Let $v = 1/y$,then $dv = -\frac{1}{y^2} dy \implies \frac{dy}{y^2} = -dv$.
Substituting these: $\int \cos(u) (2 du) = \int 2 e^v (-dv)$.
$2 \sin(u) = -2 e^v + C$.
$\sin \sqrt{x} = -e^{1/y} + C' \implies \sin \sqrt{x} + e^{1/y} = C'$.
The curve passes through $(\pi^2, 1)$,so $\sin \sqrt{\pi^2} + e^{1/1} = C' \implies \sin \pi + e = C' \implies 0 + e = C'$.
Thus,$C' = e$.
The equation is $\sin \sqrt{x} + e^{1/y} = e$.

(B) Given the differential equation $\frac{dy}{dx} = \sqrt{1-y^2}$.
Separating the variables,we get $\frac{dy}{\sqrt{1-y^2}} = dx$.
Integrating both sides,we have $\int \frac{dy}{\sqrt{1-y^2}} = \int dx$.
This results in $\sin^{-1}(y) = x + C$.
Using the initial condition $y(0) = 1$,we substitute $x = 0$ and $y = 1$ into the equation:
$\sin^{-1}(1) = 0 + C$,which gives $C = \sin^{-1}(1)$.
Substituting the value of $C$ back into the general solution,we get $\sin^{-1}(y) = x + \sin^{-1}(1)$.

(C) Given the differential equation $\frac{dy}{dx} = (3x + y + 4)^2$.
Let $3x + y + 4 = t$.
Differentiating with respect to $x$,we get $3 + \frac{dy}{dx} = \frac{dt}{dx}$,which implies $\frac{dy}{dx} = \frac{dt}{dx} - 3$.
Substituting this into the differential equation: $\frac{dt}{dx} - 3 = t^2$.
$\frac{dt}{dx} = t^2 + 3$.
Separating the variables: $\frac{dt}{t^2 + 3} = dx$.
Integrating both sides: $\int \frac{dt}{t^2 + (\sqrt{3})^2} = \int dx$.
Using the formula $\int \frac{dt}{t^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{t}{a}) + C$,we get $\frac{1}{\sqrt{3}} \tan^{-1}(\frac{t}{\sqrt{3}}) = x + k$.
Substituting $t = 3x + y + 4$ back: $\frac{1}{\sqrt{3}} \tan^{-1}(\frac{3x + y + 4}{\sqrt{3}}) - x = k$.
Comparing this with the given form $\frac{1}{\sqrt{3}} \tan^{-1}(f(x, y)) - x = k$,we identify $f(x, y) = \frac{3x + y + 4}{\sqrt{3}}$.
Now,calculate $f(1, 2) = \frac{3(1) + 2 + 4}{\sqrt{3}} = \frac{9}{\sqrt{3}} = 3\sqrt{3}$.

(A) Given,$\frac{dy}{dx} = \frac{x^3(y^4+1)}{\left[2y^{-2/3} + 3x^2y^{-2/3}\right]^{3/2}} = \frac{x^3(y^4+1)}{\left[y^{-2/3}(2+3x^2)\right]^{3/2}} = \frac{x^3(y^4+1)}{y^{-1}(2+3x^2)^{3/2}} = \frac{x^3 y(y^4+1)}{(2+3x^2)^{3/2}}$.
Separating the variables,we get $\int \frac{1}{y(y^4+1)} dy = \int \frac{x^3}{(2+3x^2)^{3/2}} dx$.
For the left side,$I_1 = \int \frac{1}{y(y^4+1)} dy = \int \frac{y^3}{y^4(y^4+1)} dy$. Let $u = y^4$,then $du = 4y^3 dy$,so $I_1 = \frac{1}{4} \int \frac{du}{u(u+1)} = \frac{1}{4} \int \left(\frac{1}{u} - \frac{1}{u+1}\right) du = \frac{1}{4} \log \left|\frac{u}{u+1}\right| = \frac{1}{4} \log \left(\frac{y^4}{1+y^4}\right)$.
For the right side,$I_2 = \int \frac{x^3}{(2+3x^2)^{3/2}} dx$. Let $t^2 = 2+3x^2$,then $2t dt = 6x dx$,so $x dx = \frac{1}{3} t dt$ and $x^2 = \frac{t^2-2}{3}$.
$I_2 = \int \frac{x^2 \cdot x dx}{(t^2)^{3/2}} = \int \frac{(\frac{t^2-2}{3}) \cdot \frac{1}{3} t dt}{t^3} = \frac{1}{9} \int \frac{t^2-2}{t^2} dt = \frac{1}{9} \int (1 - 2t^{-2}) dt = \frac{1}{9} (t + \frac{2}{t}) = \frac{1}{9} \left(\frac{t^2+2}{t}\right) = \frac{1}{9} \left(\frac{2+3x^2+2}{\sqrt{2+3x^2}}\right) = \frac{1}{9} \left(\frac{4+3x^2}{\sqrt{2+3x^2}}\right)$.
Equating $I_1 = I_2 + C$,we get $\frac{1}{4} \log \left(\frac{y^4}{1+y^4}\right) = \frac{1}{9} \left(\frac{4+3x^2}{\sqrt{2+3x^2}}\right) + C'$.
Multiplying by $4$,$\log \left(\frac{y^4}{1+y^4}\right) = \frac{4}{9} \left(\frac{4+3x^2}{\sqrt{2+3x^2}}\right) + C$.

(C) Given differential equation: $(x-2y+1)dy - (3x-6y+2)dx = 0$
$\Rightarrow \frac{dy}{dx} = \frac{3x-6y+2}{x-2y+1} = \frac{3(x-2y)+2}{(x-2y)+1}$
Let $v = x-2y$. Then $\frac{dv}{dx} = 1 - 2\frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{1}{2}(1 - \frac{dv}{dx})$.
Substituting this into the equation: $\frac{1}{2}(1 - \frac{dv}{dx}) = \frac{3v+2}{v+1}$
$1 - \frac{dv}{dx} = \frac{6v+4}{v+1} \Rightarrow \frac{dv}{dx} = 1 - \frac{6v+4}{v+1} = \frac{v+1-6v-4}{v+1} = \frac{-5v-3}{v+1}$
$\int \frac{v+1}{-5v-3} dv = \int dx$
$-\frac{1}{5} \int \frac{5v+5}{5v+3} dv = x + C_1$
$-\frac{1}{5} \int \frac{(5v+3)+2}{5v+3} dv = x + C_1$
$-\frac{1}{5} [v + \frac{2}{5} \ln|5v+3|] = x + C_1$
Substitute $v = x-2y$: $-\frac{1}{5}(x-2y) - \frac{2}{25} \ln|5(x-2y)+3| = x + C_1$
$-\frac{2}{25} \ln|5x-10y+3| = x + \frac{1}{5}x - \frac{2}{5}y + C_1 = \frac{6}{5}x - \frac{2}{5}y + C_1$
$\ln|5x-10y+3|^{-2/25} = \frac{6x-2y}{5} + C_1$
$|5x-10y+3|^{-2/25} = e^{\frac{6x-2y}{5}} \cdot e^{C_1}$
$|5(x-2y+\frac{3}{5})|^{-2/25} = e^{\frac{6x-2y}{5}} \cdot C_2$
$|x-2y+\frac{3}{5}|^{-2/25} = e^{\frac{6x-2y}{5}} \cdot C_3$
$\left|x-2y+\frac{3}{5}\right|^{2/25} \cdot e^{\frac{1}{5}(6x-2y)} = C$

(B) Given the differential equation: $(2x - 3y + 5)dx + (9y - 6x - 7)dy = 0$
Rearranging the terms: $\frac{dy}{dx} = -\frac{2x - 3y + 5}{-(6x - 9y + 7)} = \frac{2x - 3y + 5}{3(2x - 3y) + 7}$
Let $2x - 3y = z$. Then $2 - 3\frac{dy}{dx} = \frac{dz}{dx}$,which implies $\frac{dy}{dx} = \frac{1}{3}(2 - \frac{dz}{dx})$.
Substituting into the equation: $\frac{1}{3}(2 - \frac{dz}{dx}) = \frac{z + 5}{3z + 7}$
$2 - \frac{dz}{dx} = \frac{3z + 15}{3z + 7}$
$\frac{dz}{dx} = 2 - \frac{3z + 15}{3z + 7} = \frac{6z + 14 - 3z - 15}{3z + 7} = \frac{3z - 1}{3z + 7}$
Separating variables: $\frac{3z + 7}{3z - 1} dz = dx$
Integrating both sides: $\int (1 + \frac{8}{3z - 1}) dz = \int dx$
$z + \frac{8}{3} \log |3z - 1| = x + c$
Multiply by $3$: $3z + 8 \log |3z - 1| = 3x + c'$
Substitute $z = 2x - 3y$: $3(2x - 3y) + 8 \log |3(2x - 3y) - 1| = 3x + c'$
$6x - 9y + 8 \log |6x - 9y - 1| = 3x + c'$
$3x - 9y + 8 \log |6x - 9y - 1| = c$.

(D) Given the differential equation: $(y \sin x + y) \frac{dy}{dx} - \cos^2 x = 0$
Separate the variables: $y(1 + \sin x) \frac{dy}{dx} = \cos^2 x$
$y \, dy = \frac{\cos^2 x}{1 + \sin x} dx$
Using the identity $\cos^2 x = 1 - \sin^2 x = (1 - \sin x)(1 + \sin x)$:
$y \, dy = \frac{(1 - \sin x)(1 + \sin x)}{1 + \sin x} dx$
$y \, dy = (1 - \sin x) dx$
Integrating both sides: $\int y \, dy = \int (1 - \sin x) dx$
$\frac{y^2}{2} = x - (-\cos x) + c$
$\frac{y^2}{2} = x + \cos x + c$
$y^2 = 2x + 2 \cos x + c$

(D) Given the differential equation: $\frac{dy}{dx} = 1 - \cos(y-x) \cot(y-x)$.
Let $v = y - x$. Then,differentiating with respect to $x$,we get $\frac{dv}{dx} = \frac{dy}{dx} - 1$,which implies $\frac{dy}{dx} = 1 + \frac{dv}{dx}$.
Substituting these into the original equation:
$1 + \frac{dv}{dx} = 1 - \cos v \cot v$
$\frac{dv}{dx} = -\cos v \left( \frac{\cos v}{\sin v} \right) = -\frac{\cos^2 v}{\sin v}$.
Separating the variables:
$-\int \frac{\sin v}{\cos^2 v} dv = \int dx$.
Using the substitution $u = \cos v$,$du = -\sin v dv$,the integral becomes:
$\int \frac{du}{u^2} = x + c$
$-\frac{1}{u} = x + c$
$-\frac{1}{\cos v} = x + c$
$-\sec v = x + c$
$x + \sec(y-x) = c$.

(A) Given the differential equation: $\frac{dy}{dx} = \frac{1}{x+y+1}$.
Let $v = x+y+1$. Then $\frac{dv}{dx} = 1 + \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$.
Substituting this into the equation: $\frac{dv}{dx} - 1 = \frac{1}{v}$.
$\frac{dv}{dx} = \frac{1}{v} + 1 = \frac{1+v}{v}$.
Separating the variables: $\frac{v}{1+v} dv = dx$.
$\left(\frac{1+v-1}{1+v}\right) dv = dx \Rightarrow \left(1 - \frac{1}{1+v}\right) dv = dx$.
Integrating both sides: $\int (1 - \frac{1}{1+v}) dv = \int dx$.
$v - \log|1+v| = x + c$.
Substitute $v = x+y+1$: $(x+y+1) - \log|x+y+2| = x + c$.
$y + 1 - \log|x+y+2| = c$.
$y = \log|x+y+2| + c - 1$.
Let $c - 1 = \log k$,then $y = \log|x+y+2| - \log k = \log\left(\frac{x+y+2}{k}\right)$.

(D) Given the differential equation: $\frac{dy}{dx} = 1 + x + y + xy$
Factorizing the right side: $\frac{dy}{dx} = (1 + x)(1 + y)$
Separating the variables: $\frac{dy}{1 + y} = (1 + x) dx$
Integrating both sides: $\int \frac{dy}{1 + y} = \int (1 + x) dx$
This gives: $\log(1 + y) = x + \frac{x^2}{2} + c$
Taking the exponential of both sides: $1 + y = e^{x + \frac{x^2}{2} + c}$
$1 + y = e^c \cdot e^{x + \frac{x^2}{2}}$
Let $k = e^c$,then $1 + y = k e^{x + \frac{x^2}{2}}$
Therefore,the general solution is: $y = k e^{x + \frac{x^2}{2}} - 1$

(C) Given differential equation is $\frac{dy}{dx} + 2x \tan(x-y) = 1$.
Let $t = x - y$. Then $\frac{dt}{dx} = 1 - \frac{dy}{dx}$,which implies $\frac{dy}{dx} = 1 - \frac{dt}{dx}$.
Substituting this into the given equation:
$1 - \frac{dt}{dx} + 2x \tan(t) = 1$
$\Rightarrow \frac{dt}{dx} = 2x \tan(t)$
$\Rightarrow \cot(t) dt = 2x dx$.
Integrating both sides:
$\int \cot(t) dt = \int 2x dx$
$\ln|\sin(t)| = x^2 + C$.
Let $C = \ln|A|$,then $\ln|\sin(t)| = x^2 + \ln|A|$.
$\ln|\sin(t)| - \ln|A| = x^2$
$\ln|\frac{\sin(t)}{A}| = x^2$
$\sin(t) = A e^{x^2}$.
Substituting $t = x - y$ back,we get $\sin(x-y) = A e^{x^2}$.

(A) The given differential equation is $\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}+\cos x=0$.
Rearranging the terms to separate the variables,we get:
$\frac{d y}{1+y} + \frac{\cos x}{2+\sin x} dx = 0$.
Integrating both sides:
$\int \frac{d y}{1+y} + \int \frac{\cos x}{2+\sin x} dx = C_0$.
Let $t = 2+\sin x$,then $dt = \cos x dx$. The integral becomes:
$\ln|1+y| + \ln|2+\sin x| = C_1$.
This simplifies to $(1+y)(2+\sin x) = C$.
Given $y(0)=1$,we substitute $x=0$ and $y=1$:
$(1+1)(2+\sin 0) = C \Rightarrow 2(2) = C \Rightarrow C=4$.
So,the particular solution is $(1+y)(2+\sin x) = 4$.
Now,find $y\left(\frac{\pi}{2}\right)$ by substituting $x=\frac{\pi}{2}$:
$(1+y(\frac{\pi}{2}))(2+\sin(\frac{\pi}{2})) = 4$.
$(1+y(\frac{\pi}{2}))(2+1) = 4$.
$3(1+y(\frac{\pi}{2})) = 4$.
$1+y(\frac{\pi}{2}) = \frac{4}{3}$.
$y(\frac{\pi}{2}) = \frac{4}{3} - 1 = \frac{1}{3}$.

(B) Given equation: $\tan y \frac{dy}{dx} = \sin(x+y) + \sin(x-y)$
Using the identity $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
$\tan y \frac{dy}{dx} = 2 \sin x \cos y$
$\frac{\sin y}{\cos y} \frac{dy}{dx} = 2 \sin x \cos y$
Rearranging the terms:
$\frac{\sin y}{\cos^2 y} dy = 2 \sin x dx$
Integrating both sides:
$\int \frac{\sin y}{\cos^2 y} dy = \int 2 \sin x dx$
Let $t = \cos y$,then $dt = -\sin y dy$,so $-\int \frac{dt}{t^2} = -2 \cos x + c$
$\frac{1}{t} = -2 \cos x + c$
Substituting $t = \cos y$:
$\sec y = -2 \cos x + c$

(A) Given the differential equation: $\frac{dy}{dx} = \frac{x-2y+1}{2(x-2y)}$.
Let $z = x-2y$. Then,differentiating with respect to $x$,we get $\frac{dz}{dx} = 1 - 2\frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{1}{2}(1 - \frac{dz}{dx})$.
Substituting these into the original equation:
$\frac{1}{2}(1 - \frac{dz}{dx}) = \frac{z+1}{2z}$.
Multiplying both sides by $2$,we get $1 - \frac{dz}{dx} = \frac{z+1}{z} = 1 + \frac{1}{z}$.
Subtracting $1$ from both sides,we get $-\frac{dz}{dx} = \frac{1}{z}$,which simplifies to $z dz = -dx$.
Integrating both sides,we get $\int z dz = \int -dx$,which results in $\frac{z^2}{2} = -x + C_1$.
Multiplying by $2$,we get $z^2 = -2x + 2C_1$,or $z^2 + 2x = C$ (where $C = 2C_1$).
Substituting $z = x-2y$ back,we get $(x-2y)^2 + 2x = C$.

(D) Given differential equation is $\frac{dy}{dx} = \frac{xy+y}{xy+x}$.
Separating the variables,we get $\frac{dy}{dx} = \frac{y(x+1)}{x(y+1)}$.
Rearranging the terms,we have $\frac{1+y}{y} dy = \frac{1+x}{x} dx$.
This can be written as $(\frac{1}{y} + 1) dy = (\frac{1}{x} + 1) dx$.
Integrating both sides,we get $\int (\frac{1}{y} + 1) dy = \int (\frac{1}{x} + 1) dx$.
$\log|y| + y = \log|x| + x + C$.
Rearranging the terms,$y - x = \log|x| - \log|y| + C$.
$y - x = \log|\frac{x}{y}| + C$.
Let $C = \log c$,then $y - x = \log|\frac{x}{y}| + \log c = \log|\frac{cx}{y}|$.
Thus,the solution is $y - x = \log \left(\frac{cx}{y}\right)$.

(C) Given that $dx + dy = (x + y)(dx - dy)$.
Dividing by $dx$,we get:
$1 + \frac{dy}{dx} = (x + y)(1 - \frac{dy}{dx})$
$1 + \frac{dy}{dx} = x + y - (x + y)\frac{dy}{dx}$
$\frac{dy}{dx}(1 + x + y) = x + y - 1$
$\frac{dy}{dx} = \frac{x + y - 1}{x + y + 1} \quad \dots(i)$
Let $x + y = t$. Then $1 + \frac{dy}{dx} = \frac{dt}{dx}$,so $\frac{dy}{dx} = \frac{dt}{dx} - 1$.
Substituting into Eq. $(i)$:
$\frac{dt}{dx} - 1 = \frac{t - 1}{t + 1}$
$\frac{dt}{dx} = \frac{t - 1}{t + 1} + 1 = \frac{t - 1 + t + 1}{t + 1} = \frac{2t}{t + 1}$
Separating variables:
$\frac{t + 1}{2t} dt = dx$
$\frac{1}{2}(1 + \frac{1}{t}) dt = dx$
Integrating both sides:
$\frac{1}{2}(t + \log|t|) = x + C_1$
$t + \log|t| = 2x + 2C_1$
Substituting $t = x + y$:
$x + y + \log(x + y) = 2x + C$
$\log(x + y) = x - y + C$

(A) Given the differential equation: $\frac{dy}{dx} = \frac{x \log x^2 + x}{\sin y + y \cos y}$.
By separating the variables,we get: $(\sin y + y \cos y) dy = (x \log x^2 + x) dx$.
Integrating both sides: $\int (\sin y + y \cos y) dy = \int (x \log x^2 + x) dx$.
For the left side,using integration by parts on $\int y \cos y dy$: $\int y \cos y dy = y \sin y - \int \sin y dy = y \sin y + \cos y$.
Thus,$\int (\sin y + y \cos y) dy = -\cos y + y \sin y + \cos y = y \sin y$.
For the right side,$\int (x \log x^2 + x) dx = \int (2x \log x + x) dx$.
Using integration by parts for $\int 2x \log x dx$: Let $u = \log x$,$dv = 2x dx$,then $du = \frac{1}{x} dx$,$v = x^2$.
$\int 2x \log x dx = x^2 \log x - \int x^2 \cdot \frac{1}{x} dx = x^2 \log x - \int x dx = x^2 \log x - \frac{x^2}{2}$.
Adding the integral of $x$: $\int (2x \log x + x) dx = x^2 \log x - \frac{x^2}{2} + \frac{x^2}{2} + C = x^2 \log x + C$.
Equating both sides: $y \sin y = x^2 \log x + C$.

(A) Given the differential equation: $x^2 + y^2 \frac{dy}{dx} = 4$.
Rearranging the terms to separate the variables,we get: $y^2 dy = (4 - x^2) dx$.
Integrating both sides: $\int y^2 dy = \int (4 - x^2) dx$.
This yields: $\frac{y^3}{3} = 4x - \frac{x^3}{3} + C_1$.
Multiplying the entire equation by $3$: $y^3 = 12x - x^3 + 3C_1$.
Let $3C_1 = C$,then the solution is: $x^3 + y^3 = 12x + C$.

(B) Given differential equation: $(9x - 3y + 5) dy = (3x - y + 1) dx$
$\frac{dy}{dx} = \frac{3x - y + 1}{3(3x - y) + 5}$
Let $v = 3x - y$. Then $\frac{dv}{dx} = 3 - \frac{dy}{dx}$,so $\frac{dy}{dx} = 3 - \frac{dv}{dx}$.
Substituting into the equation: $3 - \frac{dv}{dx} = \frac{v + 1}{3v + 5}$
$\frac{dv}{dx} = 3 - \frac{v + 1}{3v + 5} = \frac{9v + 15 - v - 1}{3v + 5} = \frac{8v + 14}{3v + 5}$
Separating variables: $\frac{3v + 5}{8v + 14} dv = dx$
$\frac{1}{8} \int \frac{3v + 5}{v + 1.75} dv = x + C$
Using partial fractions or adjustment: $\frac{3v + 5}{8v + 14} = \frac{3}{8} \left( \frac{8v + 14 - 14 + 13.33}{8v + 14} \right) = \frac{3}{8} \left( 1 + \frac{-14/3 + 5}{8v + 14} \right) = \frac{3}{8} - \frac{1}{8(8v + 14)} \dots$
Integrating: $\frac{3}{8} v - \frac{1}{64} \ln |8v + 14| = x + C$
Substitute $v = 3x - y$: $\frac{3}{8}(3x - y) - \frac{1}{64} \ln |8(3x - y) + 14| = x + C$
Multiply by $64$: $24(3x - y) - \ln |24x - 8y + 14| = 64x + C'$
$72x - 24y - \ln |2(12x - 4y + 7)| = 64x + C'$
$8x - 24y - \ln |12x - 4y + 7| = C''$
Dividing by $2$: $4x - 12y - \frac{1}{2} \ln |12x - 4y + 7| = C$. This matches option $B$.

(B) Given the differential equation: $\frac{dy}{dx} = \frac{x+y-3}{x+y-7}$.
Let $v = x+y$. Then $\frac{dv}{dx} = 1 + \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$.
Substituting this into the equation: $\frac{dv}{dx} - 1 = \frac{v-3}{v-7}$.
$\frac{dv}{dx} = \frac{v-3}{v-7} + 1 = \frac{v-3+v-7}{v-7} = \frac{2v-10}{v-7} = \frac{2(v-5)}{v-7}$.
Separating the variables: $\int \frac{v-7}{v-5} dv = \int 2 dx$.
$\int \frac{(v-5)-2}{v-5} dv = 2x + C$.
$\int (1 - \frac{2}{v-5}) dv = 2x + C$.
$v - 2 \ln |v-5| = 2x + C$.
Substitute $v = x+y$: $(x+y) - 2 \ln |x+y-5| = 2x + C$.
$y-x-C = 2 \ln |x+y-5|$.
$\ln |x+y-5|^2 = y-x-C$.
Taking the exponential of both sides: $(x+y-5)^2 = e^{y-x-C} = C' e^{y-x}$.
Thus,the correct option is $B$.

(C) We have,$\frac{dy}{dx} = (4x + y + 1)^2$.
Let $v = 4x + y + 1$.
Then,$\frac{dv}{dx} = 4 + \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 4$.
Substituting this into the differential equation,we get $\frac{dv}{dx} - 4 = v^2$,or $\frac{dv}{dx} = v^2 + 4$.
Separating the variables,we have $\frac{dv}{v^2 + 4} = dx$.
Integrating both sides,$\int \frac{dv}{v^2 + 2^2} = \int dx$.
This gives $\frac{1}{2} \tan^{-1}(\frac{v}{2}) = x + c$.
Substituting $v = 4x + y + 1$,we get $\frac{1}{2} \tan^{-1}(\frac{4x + y + 1}{2}) = x + c$.
Given $y(0) = 1$,we substitute $x = 0$ and $y = 1$: $\frac{1}{2} \tan^{-1}(\frac{0 + 1 + 1}{2}) = 0 + c$,so $c = \frac{1}{2} \tan^{-1}(1) = \frac{\pi}{8}$.
Thus,$\frac{1}{2} \tan^{-1}(\frac{4x + y + 1}{2}) = x + \frac{\pi}{8}$.
Multiplying by $2$,$\tan^{-1}(\frac{4x + y + 1}{2}) = 2x + \frac{\pi}{4}$.
Taking the tangent of both sides,$\frac{4x + y + 1}{2} = \tan(2x + \frac{\pi}{4})$.
Therefore,$y = 2 \tan(2x + \frac{\pi}{4}) - 4x - 1$.

(C) Given differential equation: $\cos (x+y) dy = dx$
Rearranging,we get: $\frac{dx}{dy} = \cos (x+y)$
Let $x+y = v$. Then,differentiating with respect to $y$,we get: $\frac{dx}{dy} + 1 = \frac{dv}{dy} \implies \frac{dx}{dy} = \frac{dv}{dy} - 1$
Substituting this into the equation: $\frac{dv}{dy} - 1 = \cos v$
$\frac{dv}{dy} = 1 + \cos v$
Separating variables: $\int \frac{dv}{1 + \cos v} = \int dy$
Using the identity $1 + \cos v = 2 \cos^2 \frac{v}{2}$:
$\int \frac{dv}{2 \cos^2 \frac{v}{2}} = \int dy$
$\frac{1}{2} \int \sec^2 \frac{v}{2} dv = \int dy$
Integrating both sides: $\tan \frac{v}{2} = y + c$
Substituting $v = x+y$ back: $\tan \frac{x+y}{2} = y + c$
Thus,$y = \tan \frac{x+y}{2} + c$.

(B) Given differential equation is $2 dx + dy = (6xy + 4x - 3y) dx$.
Rearranging the terms: $dy = (6xy + 4x - 3y - 2) dx$.
$dy = [2x(3y + 2) - (3y + 2)] dx$.
$dy = (2x - 1)(3y + 2) dx$.
Separating the variables: $\frac{dy}{3y + 2} = (2x - 1) dx$.
Integrating both sides: $\int \frac{1}{3y + 2} dy = \int (2x - 1) dx$.
$\frac{1}{3} \log |3y + 2| = x^2 - x + C_1$.
Multiplying by $3$: $\log |3y + 2| = 3x^2 - 3x + 3C_1$.
Let $3C_1 = c$,then $\log |3y + 2| = 3x^2 - 3x + c$.
Thus,the correct option is $B$.

(B) Given differential equation is $\left(\left(1+x^2\right) y \sin x-2 x y\right) d x-\log y^{1+x^2} d y=0$.
Dividing by $y$ and using $\log y^{1+x^2} = (1+x^2) \log y$,we get:
$\left((1+x^2) \sin x - 2x\right) dx - (1+x^2) \log y \frac{dy}{y} = 0$.
Rearranging the terms:
$\left(\sin x - \frac{2x}{1+x^2}\right) dx = \frac{\log y}{y} dy$.
Integrating both sides:
$\int \left(\sin x - \frac{2x}{1+x^2}\right) dx = \int \frac{\log y}{y} dy$.
Let $u = \log y$,then $du = \frac{1}{y} dy$.
Integrating gives:
$-\cos x - \log(1+x^2) = \frac{(\log y)^2}{2} + C_1$.
Multiplying by $2$:
$-2 \cos x - 2 \log(1+x^2) = (\log y)^2 + 2C_1$.
Rearranging:
$(\log y)^2 + 2 \cos x + 2 \log(1+x^2) = C$.
Since $2 \log(1+x^2) = \log(1+x^2)^2$,the solution is $(\log y)^2 + 2 \cos x + \log(1+x^2)^2 = C$.

(B) Given the differential equation: $\frac{dy}{dx} = x + \sin x \cos y + x \cos y + \sin x$
Factor the right side: $\frac{dy}{dx} = x(1 + \cos y) + \sin x(1 + \cos y)$
$\frac{dy}{dx} = (x + \sin x)(1 + \cos y)$
Separate the variables: $\frac{dy}{1 + \cos y} = (x + \sin x) dx$
Using the identity $1 + \cos y = 2 \cos^2 \frac{y}{2}$,we get: $\frac{dy}{2 \cos^2 \frac{y}{2}} = (x + \sin x) dx$
$\frac{1}{2} \sec^2 \frac{y}{2} dy = (x + \sin x) dx$
Integrating both sides: $\int \frac{1}{2} \sec^2 \frac{y}{2} dy = \int (x + \sin x) dx$
$\tan \frac{y}{2} = \frac{x^2}{2} - \cos x + C$

(A) Given the differential equation: $x dx + y dy = x^2 y dy - x y^2 dx$
Rearranging the terms to separate variables:
$x dx + x y^2 dx = x^2 y dy - y dy$
$x(1 + y^2) dx = y(x^2 - 1) dy$
Separating the variables:
$\frac{x}{x^2 - 1} dx = \frac{y}{1 + y^2} dy$
Multiplying both sides by $2$:
$\frac{2x}{x^2 - 1} dx = \frac{2y}{1 + y^2} dy$
Integrating both sides:
$\int \frac{2x}{x^2 - 1} dx = \int \frac{2y}{1 + y^2} dy$
$\ln|x^2 - 1| = \ln|1 + y^2| + \ln C$
Using the property $\ln a + \ln b = \ln(ab)$:
$\ln|x^2 - 1| = \ln|C(1 + y^2)|$
Taking the exponential of both sides:
$x^2 - 1 = C(1 + y^2)$

(B) Given the differential equation: $\frac{dy}{dx} = \sin(x-y) + \cos(x-y)$.
Let $x-y = v$. Then $1 - \frac{dy}{dx} = \frac{dv}{dx}$,which implies $\frac{dy}{dx} = 1 - \frac{dv}{dx}$.
Substituting this into the equation: $1 - \frac{dv}{dx} = \sin v + \cos v$.
Rearranging gives: $\frac{dv}{dx} = 1 - (\sin v + \cos v)$,so $\frac{dv}{1 - (\sin v + \cos v)} = dx$.
Using half-angle formulas $\sin v = \frac{2 \tan(v/2)}{1 + \tan^2(v/2)}$ and $\cos v = \frac{1 - \tan^2(v/2)}{1 + \tan^2(v/2)}$:
$\frac{dv}{1 - \left(\frac{2 \tan(v/2) + 1 - \tan^2(v/2)}{1 + \tan^2(v/2)}\right)} = dx$.
Simplifying the denominator: $\frac{(1 + \tan^2(v/2)) dv}{1 + \tan^2(v/2) - 2 \tan(v/2) - 1 + \tan^2(v/2)} = dx$.
This simplifies to $\frac{\sec^2(v/2) dv}{2 \tan^2(v/2) - 2 \tan(v/2)} = dx$.
Let $u = \tan(v/2)$,then $du = \frac{1}{2} \sec^2(v/2) dv$,so $\sec^2(v/2) dv = 2 du$.
Substituting: $\frac{2 du}{2(u^2 - u)} = dx \Rightarrow \frac{du}{u(u-1)} = dx$.
Using partial fractions: $\int (\frac{1}{u-1} - \frac{1}{u}) du = \int dx$.
Integrating gives $\log|u-1| - \log|u| = x + C$.
Substituting $u = \tan(\frac{x-y}{2})$: $\log \left| \frac{\tan(\frac{x-y}{2}) - 1}{\tan(\frac{x-y}{2})} \right| = x + C$.