Linear differential equations Questions in English

(D) The given differential equation is $\frac{dy}{dx}+y \tan x=2x+x^2 \tan x$.
This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$,where $P=\tan x$ and $Q=2x+x^2 \tan x$.
The integrating factor $IF = e^{\int \tan x dx} = e^{\ln |\sec x|} = \sec x$.
The solution is given by $y \cdot IF = \int Q \cdot IF dx + C$.
$y \sec x = \int (2x+x^2 \tan x) \sec x dx + C$.
$y \sec x = \int 2x \sec x dx + \int x^2 \sec x \tan x dx + C$.
Using integration by parts on the second integral: $\int x^2 \sec x \tan x dx = x^2 \sec x - \int 2x \sec x dx$.
Substituting this back: $y \sec x = \int 2x \sec x dx + x^2 \sec x - \int 2x \sec x dx + C$.
$y \sec x = x^2 \sec x + C$,which simplifies to $y = x^2 + C \cos x$.
Given $Y(0)=1$,we have $1 = 0^2 + C \cos(0) \implies C=1$.
So,$Y(x) = x^2 + \cos x$.
Then $Y'(x) = 2x - \sin x$.
Calculating $Y'(\frac{\pi}{4}) - Y'(-\frac{\pi}{4}) = (2(\frac{\pi}{4}) - \sin(\frac{\pi}{4})) - (2(-\frac{\pi}{4}) - \sin(-\frac{\pi}{4}))$.
$= (\frac{\pi}{2} - \frac{1}{\sqrt{2}}) - (-\frac{\pi}{2} + \frac{1}{\sqrt{2}}) = \pi - \frac{2}{\sqrt{2}} = \pi - \sqrt{2}$.
Thus,option $D$ is correct.

(C) Given differential equation: $(e^{y-x}) dy = (e^x - e^y) dx$
Multiply both sides by $e^x$: $e^y dy = (e^{2x} - e^x e^y) dx$
Rearrange the terms: $e^y \frac{dy}{dx} + e^x e^y = e^{2x}$
Let $z = e^y$,then $\frac{dz}{dx} = e^y \frac{dy}{dx}$.
The equation becomes: $\frac{dz}{dx} + e^x z = e^{2x}$.
This is a linear differential equation of the form $\frac{dz}{dx} + P(x)z = Q(x)$,where $P(x) = e^x$ and $Q(x) = e^{2x}$.
Integrating factor $IF = e^{\int P(x) dx} = e^{\int e^x dx} = e^{e^x}$.
The solution is $z \cdot IF = \int Q(x) \cdot IF dx + c$.
$z \cdot e^{e^x} = \int e^{2x} \cdot e^{e^x} dx + c$.
Let $u = e^x$,then $du = e^x dx$.
$z \cdot e^{e^x} = \int u \cdot e^u du + c$.
Using integration by parts: $\int u e^u du = u e^u - e^u + c$.
Substituting back $z = e^y$ and $u = e^x$: $e^y e^{e^x} = e^x e^{e^x} - e^{e^x} + c$.

(A) The given linear differential equation is $x \frac{dy}{dx} + 3y = x^2$.
Dividing both sides by $x$,we get:
$\frac{dy}{dx} + \frac{3}{x} y = x$.
Comparing this with the standard form $\frac{dy}{dx} + Py = Q$,we identify $P = \frac{3}{x}$.
The integrating factor $(IF)$ is given by $IF = e^{\int P dx}$.
$IF = e^{\int \frac{3}{x} dx} = e^{3 \ln x} = e^{\ln x^3} = x^3$.
Thus,the integrating factor is $x^3$.

(B) The given differential equation is $\frac{dy}{dx} + g'(x)y = g(x)g'(x)$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = g'(x)$ and $Q(x) = g(x)g'(x)$.
The integrating factor $(IF)$ is given by $IF = e^{\int P(x) dx} = e^{\int g'(x) dx} = e^{g(x)}$.
The general solution is $y(IF) = \int Q(IF) dx + C$.
Substituting the values,we get $y e^{g(x)} = \int g(x)g'(x) e^{g(x)} dx + C$.
Let $g(x) = t$,then $g'(x) dx = dt$.
So,$y e^{g(x)} = \int t e^t dt + C$.
Using integration by parts,$\int t e^t dt = t e^t - e^t + C = e^t(t - 1) + C$.
Thus,$y e^{g(x)} = e^{g(x)}(g(x) - 1) + C$.
Rearranging the terms,$y e^{g(x)} - e^{g(x)}(g(x) - 1) = C$.
$e^{g(x)}(y - g(x) + 1) = C$.
Taking the natural logarithm on both sides,$\log(e^{g(x)}) + \log|y - g(x) + 1| = \log(C)$.
$g(x) + \log|y - g(x) + 1| = C'$.
Thus,option $(b)$ is correct.

(D) The given differential equation is:
$\frac{dy}{dx} + \csc^2 (x) \cdot y = \csc^2 (x) \cdot \cot x . . . . . . . . . . (1)$
Comparing this with the linear differential equation form $\frac{dy}{dx} + Py = Q$,we get:
$P = \csc^2 x$
$Q = \csc^2 x \cdot \cot x$
Now,find the Integrating Factor $(IF)$:
$IF = e^{\int P \ dx} = e^{\int \csc^2 x \ dx} = e^{- \cot x}$
The general solution is given by:
$y \cdot IF = \int Q \cdot IF \ dx + C$
$y \cdot e^{- \cot x} = \int \csc^2 x \cdot \cot x \cdot e^{- \cot x} \ dx + C$
Let $t = \cot x$,then $dt = - \csc^2 x \ dx$,which implies $\csc^2 x \ dx = - dt$.
Substituting these into the integral:
$y \cdot e^{- \cot x} = \int t \cdot e^{- t} (- dt) = - \int t e^{- t} \ dt$
Using integration by parts $\int u \ dv = uv - \int v \ du$ where $u = t$ and $dv = e^{- t} \ dt$:
$- \int t e^{- t} \ dt = - [t (- e^{- t}) - \int (- e^{- t}) \ dt] = - [- t e^{- t} - e^{- t}] + C = t e^{- t} + e^{- t} + C$
Substituting $t = \cot x$ back:
$y \cdot e^{- \cot x} = e^{- \cot x} (\cot x + 1) + C$.

(A) The given differential equation is $\frac{dy}{dx} + 2y \tan x = \sin x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 2 \tan x$ and $Q = \sin x$.
The integrating factor $(I.F.)$ is $e^{\int P dx} = e^{\int 2 \tan x dx} = e^{2 \ln |\sec x|} = e^{\ln \sec^2 x} = \sec^2 x$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
$y \sec^2 x = \int \sin x \cdot \sec^2 x dx + c$.
$y \sec^2 x = \int \sec x \tan x dx + c$.
$y \sec^2 x = \sec x + c$.
Given $y = 0$ when $x = \frac{\pi}{3}$,we substitute these values:
$0 \cdot \sec^2(\frac{\pi}{3}) = \sec(\frac{\pi}{3}) + c$.
$0 = 2 + c \implies c = -2$.
Thus,$y \sec^2 x = \sec x - 2$.
Dividing by $\sec^2 x$,we get $y = \frac{\sec x}{\sec^2 x} - \frac{2}{\sec^2 x} = \cos x - 2 \cos^2 x$.
Using $\cos^2 x = 1 - \sin^2 x$,we get $y = \cos x - 2(1 - \sin^2 x) = \cos x - 2 + 2 \sin^2 x$.
Therefore,$y = 2 \sin^2 x + \cos x - 2$.

(D) Given the linear differential equation: $\frac{dy}{dx} + P(x) y = Q(x)$.
The integrating factor $(IF)$ is $e^{\int P(x) dx}$.
The general solution is $y \cdot IF = \int Q(x) \cdot IF dx + C$.
Given $y = A(x) e^{\int P(x) dx}$,we substitute this into the differential equation:
$\frac{d}{dx} [A(x) e^{\int P(x) dx}] + P(x) [A(x) e^{\int P(x) dx}] = Q(x)$.
Using the product rule:
$A'(x) e^{\int P(x) dx} + A(x) P(x) e^{\int P(x) dx} + P(x) A(x) e^{\int P(x) dx} = Q(x)$.
$A'(x) e^{\int P(x) dx} + 2 A(x) P(x) e^{\int P(x) dx} = Q(x)$.
Wait,let us use the standard form $y = u v$ where $v = e^{\int P dx}$.
Then $y' = u' v + u v' = u' e^{\int P dx} + u P e^{\int P dx}$.
Substituting into $\frac{dy}{dx} + P y = Q$:
$u' e^{\int P dx} + u P e^{\int P dx} + P u e^{\int P dx} = Q$ is incorrect.
Correct approach: $y = A(x) e^{\int P dx}$.
$\frac{dy}{dx} = A'(x) e^{\int P dx} + A(x) P(x) e^{\int P dx}$.
Substitute into $\frac{dy}{dx} + P y = Q$:
$A'(x) e^{\int P dx} + A(x) P(x) e^{\int P dx} + P(x) A(x) e^{\int P dx} = Q(x)$ is not the standard method.
Actually,the solution is $y e^{\int P dx} = \int Q e^{\int P dx} dx + C$.
Comparing $y = A(x) e^{\int P dx}$,we have $A(x) = \int Q e^{\int P dx} dx$.
Therefore,$A'(x) = \frac{d}{dx} [\int Q e^{\int P dx} dx] = Q(x) e^{\int P dx}$.

(B) Given differential equation is $\sin y \frac{dy}{dx} = \cos y(1 - x \cos y)$.
Dividing both sides by $\cos^2 y$,we get:
$\frac{\sin y}{\cos^2 y} \frac{dy}{dx} = \frac{1}{\cos y} - x$
$\sec y \tan y \frac{dy}{dx} = \sec y - x$
Let $\sec y = t$. Then $\sec y \tan y \frac{dy}{dx} = \frac{dt}{dx}$.
Substituting this into the equation,we get the linear differential equation:
$\frac{dt}{dx} = t - x \implies \frac{dt}{dx} - t = -x$.
The integrating factor $(IF)$ is $e^{\int -1 dx} = e^{-x}$.
The solution is $t(IF) = \int (-x)(IF) dx + c$.
$t e^{-x} = \int -x e^{-x} dx + c$.
Using integration by parts for $\int -x e^{-x} dx$:
$\int -x e^{-x} dx = x e^{-x} - \int e^{-x} dx = x e^{-x} + e^{-x} + c$.
Thus,$t e^{-x} = x e^{-x} + e^{-x} + c$.
Multiplying by $e^x$,we get $t = x + 1 + c e^x$.
Substituting back $t = \sec y$,we get $\sec y = x + 1 + c e^x$.

(D) The given differential equation is $\frac{dy}{dx} + y \tan x = 2x + x^2 \tan x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \tan x$ and $Q = 2x + x^2 \tan x$.
First,we find the Integrating Factor ($I$.$F$.):
$I.F. = e^{\int P dx} = e^{\int \tan x dx} = e^{\ln |\sec x|} = \sec x$.
The general solution is given by:
$y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$
$y \sec x = \int (2x + x^2 \tan x) \sec x dx + c$
$y \sec x = \int 2x \sec x dx + \int x^2 \tan x \sec x dx + c$
Using integration by parts for the second integral $\int x^2 (\tan x \sec x) dx$:
Let $u = x^2$ and $dv = \sec x \tan x dx$. Then $du = 2x dx$ and $v = \sec x$.
$\int x^2 \sec x \tan x dx = x^2 \sec x - \int 2x \sec x dx$.
Substituting this back into the equation:
$y \sec x = \int 2x \sec x dx + (x^2 \sec x - \int 2x \sec x dx) + c$
$y \sec x = x^2 \sec x + c$
Multiplying both sides by $\cos x$:
$y = x^2 + c \cos x$.

(C) Given differential equation is $\frac{dx}{dy} + \frac{x}{y} = x^2$.
Dividing both sides by $x^2$,we get $\frac{1}{x^2} \frac{dx}{dy} + \frac{1}{xy} = 1$.
Let $t = \frac{1}{x}$,then $\frac{dt}{dy} = -\frac{1}{x^2} \frac{dx}{dy}$.
Substituting this into the equation,we get $-\frac{dt}{dy} + \frac{t}{y} = 1$,which simplifies to $\frac{dt}{dy} - \frac{t}{y} = -1$.
This is a linear differential equation of the form $\frac{dt}{dy} + P(y)t = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = -1$.
The integrating factor $I.F. = e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\log y} = \frac{1}{y}$.
The solution is $t \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$.
$t \cdot \frac{1}{y} = \int (-1) \cdot \frac{1}{y} dy + c$.
$\frac{t}{y} = -\log |y| + c$.
Substituting $t = \frac{1}{x}$,we get $\frac{1}{xy} = -\log |y| + c$,or $\frac{1}{x} = cy - y \log |y|$.

(C) Given,$\cos^{2} x \cdot \frac{dy}{dx} - (\tan 2x) y = \cos^{4} x$
Dividing by $\cos^{2} x$,we get:
$\frac{dy}{dx} - \left(\frac{\tan 2x}{\cos^{2} x}\right) y = \cos^{2} x$
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\frac{\tan 2x}{\cos^{2} x}$ and $Q = \cos^{2} x$.
The Integrating Factor $(I.F.)$ is $e^{\int P dx}$.
$\int P dx = -\int \frac{\tan 2x}{\cos^{2} x} dx = -\int \frac{2 \tan x}{(1 - \tan^{2} x) \cos^{2} x} dx$.
Let $\tan x = t$,then $\sec^{2} x dx = dt$.
$\int P dx = -\int \frac{2t}{1 - t^{2}} dt = \ln |1 - t^{2}| = \ln |1 - \tan^{2} x|$.
Thus,$I.F. = e^{\ln |1 - \tan^{2} x|} = 1 - \tan^{2} x$.
The general solution is $y \cdot I.F. = \int (Q \cdot I.F.) dx + C$.
$y(1 - \tan^{2} x) = \int (\cos^{2} x (1 - \tan^{2} x)) dx + C$.
$y(1 - \tan^{2} x) = \int (\cos^{2} x - \sin^{2} x) dx + C$.
$y(1 - \tan^{2} x) = \int \cos 2x dx + C$.
$y(1 - \tan^{2} x) = \frac{\sin 2x}{2} + C$.
$y = \frac{1}{2} \left[ \frac{\sin 2x + 2C}{1 - \tan^{2} x} \right]$.
Replacing $2C$ with $c$,we get $y = \frac{1}{2} \left[ \frac{\sin 2x + c}{1 - \tan^{2} x} \right]$.

(D) Given differential equation is $(x+1) \frac{dy}{dx} - xy = 1$.
Dividing by $(x+1)$,we get $\frac{dy}{dx} - \frac{x}{x+1}y = \frac{1}{x+1}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\frac{x}{x+1} = -(\frac{x+1-1}{x+1}) = -1 + \frac{1}{x+1}$ and $Q = \frac{1}{x+1}$.
The integrating factor $IF = e^{\int P dx} = e^{\int (-1 + \frac{1}{x+1}) dx} = e^{-x + \log(x+1)} = e^{-x} \cdot (x+1)$.
The general solution is $y \cdot IF = \int (Q \cdot IF) dx + C$.
$y \cdot (x+1)e^{-x} = \int (\frac{1}{x+1} \cdot (x+1)e^{-x}) dx + C$.
$y(x+1)e^{-x} = \int e^{-x} dx + C = -e^{-x} + C$.
Dividing by $e^{-x}$,we get $y(x+1) = -1 + Ce^x$.
Given $y(0) = 1$,substituting $x=0$ and $y=1$: $1(0+1) = -1 + Ce^0 \implies 1 = -1 + C \implies C = 2$.
Thus,$y(x+1) = -1 + 2e^x$,which gives $y = \frac{2e^x - 1}{x+1}$.

(B) Given the differential equation: $(x-4y^3) \frac{dy}{dx}-y=0$ for $y>0$.
Rearranging the terms,we get: $(x-4y^3) \frac{dy}{dx}=y$.
Since $y>0$,we can write: $\frac{dx}{dy} = \frac{x-4y^3}{y} = \frac{x}{y} - 4y^2$.
This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = -\frac{1}{y}$ and $Q = -4y^2$.
The integrating factor ($I$.$F$.) is given by: $I.F. = e^{\int P dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln y} = e^{\ln y^{-1}} = \frac{1}{y}$.
The general solution is: $x \cdot (I.F.) = \int (Q \cdot I.F.) dy + C$.
Substituting the values: $x \cdot \frac{1}{y} = \int (-4y^2 \cdot \frac{1}{y}) dy + C$.
$\frac{x}{y} = \int -4y dy + C$.
$\frac{x}{y} = -4 \cdot \frac{y^2}{2} + C = -2y^2 + C$.
Multiplying by $y$,we get: $x = -2y^3 + Cy$,which simplifies to $x+2y^3=Cy$.

(C) Given the differential equation: $(y^3 + x) \frac{dy}{dx} = y$.
Rearranging the equation: $\frac{dx}{dy} = \frac{y^3 + x}{y} = y^2 + \frac{x}{y}$.
This is a linear differential equation in $x$: $\frac{dx}{dy} - \frac{1}{y}x = y^2$.
The integrating factor is $IF = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = \frac{1}{y}$.
Multiplying by $IF$: $\frac{1}{y} \frac{dx}{dy} - \frac{1}{y^2} x = y$.
Integrating both sides with respect to $y$: $\int \frac{d}{dy} (\frac{x}{y}) dy = \int y dy$.
$\frac{x}{y} = \frac{y^2}{2} + C$.
$x = \frac{y^3}{2} + Cy \implies 2x = y^3 + 2Cy$.
Since $y^3 = ax + b$,we rewrite as $y^3 = 2x - 2Cy$.
Comparing $y^3 = ax + b$,we see $a = 2$ and $b = -2Cy$.
Using $y(4) = 2$: $2^3 = 2(4) + b \implies 8 = 8 + b \implies b = 0$.
Then $y^3 = 2x + 0$,so $a = 2$ and $b = 0$.
Calculating $4a + 12b^2 = 4(2) + 12(0)^2 = 8 + 0 = 8$.

(C) Given differential equation: $(1+y^2) + (x - e^{\tan^{-1} y}) \frac{dx}{dy} = 0$.
Rearranging the terms: $(x - e^{\tan^{-1} y}) \frac{dx}{dy} = -(1+y^2)$,which implies $\frac{dx}{dy} = -\frac{1+y^2}{x - e^{\tan^{-1} y}}$.
This is not standard. Let us rewrite as: $\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{e^{\tan^{-1} y}}{1+y^2}$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{1+y^2}$ and $Q(y) = \frac{e^{\tan^{-1} y}}{1+y^2}$.
Integrating Factor $(IF) = e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy} = e^{\tan^{-1} y}$.
The general solution is $x \cdot IF = \int Q(y) \cdot IF dy + C$.
Substituting the values: $x e^{\tan^{-1} y} = \int \frac{e^{\tan^{-1} y}}{1+y^2} \cdot e^{\tan^{-1} y} dy + C$.
$x e^{\tan^{-1} y} = \int \frac{e^{2 \tan^{-1} y}}{1+y^2} dy + C$.
Let $u = \tan^{-1} y$,then $du = \frac{1}{1+y^2} dy$.
$x e^{\tan^{-1} y} = \int e^{2u} du + C = \frac{1}{2} e^{2u} + C = \frac{1}{2} e^{2 \tan^{-1} y} + C$.
Multiplying by $2$: $2 x e^{\tan^{-1} y} = e^{2 \tan^{-1} y} + C$.

(A) Given differential equation is $\frac{d y}{d x}+\frac{1}{x}=\frac{e^y}{x^2}$.
Dividing both sides by $e^y$,we get $e^{-y} \frac{d y}{d x} + \frac{1}{x} e^{-y} = \frac{1}{x^2} \quad \dots (i)$.
Let $e^{-y} = v$. Then,differentiating with respect to $x$,we get $-e^{-y} \frac{d y}{d x} = \frac{d v}{d x}$,or $e^{-y} \frac{d y}{d x} = -\frac{d v}{d x}$.
Substituting this into equation $(i)$,we get $-\frac{d v}{d x} + \frac{1}{x} v = \frac{1}{x^2}$,which simplifies to $\frac{d v}{d x} - \frac{1}{x} v = -\frac{1}{x^2}$.
This is a linear differential equation of the form $\frac{d v}{d x} + P(x) v = Q(x)$,where $P(x) = -\frac{1}{x}$ and $Q(x) = -\frac{1}{x^2}$.
The integrating factor $IF = e^{\int P(x) d x} = e^{\int -\frac{1}{x} d x} = e^{-\log x} = e^{\log(x^{-1})} = \frac{1}{x}$.
The solution is given by $v \cdot (IF) = \int Q(x) \cdot (IF) d x + C$.
$v \cdot \frac{1}{x} = \int \left(-\frac{1}{x^2}\right) \cdot \frac{1}{x} d x + C$.
$\frac{v}{x} = -\int x^{-3} d x + C = -\left(\frac{x^{-2}}{-2}\right) + C = \frac{1}{2 x^2} + C$.
Multiplying by $2x^2$,we get $2 x v = 1 + 2 C x^2$. Let $2C = C'$,then $2 x v = 1 + C' x^2$.
Substituting $v = e^{-y}$,we get $2 x e^{-y} = 1 + C' x^2$.
Multiplying by $e^y$,we get $2 x = (1 + C' x^2) e^y$.

(B) Given the differential equation: $\frac{dy}{dx} = \frac{1}{ax + by + c}$.
Taking the reciprocal of both sides,we get: $\frac{dx}{dy} = ax + by + c$.
Rearranging the terms,we have: $\frac{dx}{dy} - ax = by + c$.
This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = -a$ and $Q = by + c$.
Since the equation is in the form of a linear differential equation in $x$,the correct option is $B$.

(D) Given differential equation is: $(1+y+x^2 y) dx + (x+x^3) dy = 0$.
Rearranging the terms,we get: $(x+x^3) dy = -(1+y+x^2 y) dx$.
Dividing by $dx(x+x^3)$,we get: $\frac{dy}{dx} = -\frac{1+y(1+x^2)}{x(1+x^2)}$.
This simplifies to: $\frac{dy}{dx} = -\frac{1}{x(1+x^2)} - \frac{y(1+x^2)}{x(1+x^2)}$.
Thus,$\frac{dy}{dx} + \frac{y}{x} = -\frac{1}{x(1+x^2)}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = -\frac{1}{x(1+x^2)}$.
The integrating factor $(IF)$ is given by $e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\log x} = x$.
Wait,let us re-evaluate the rearrangement: $(1+y+x^2 y) dx + x(1+x^2) dy = 0$.
Dividing by $dx(1+x^2)$,we get $\frac{1+y(1+x^2)}{1+x^2} dx + x dy = 0$.
This is not standard. Let us rewrite as $\frac{dy}{dx} + \frac{y}{x} = -\frac{1}{x(1+x^2)}$.
Actually,the integrating factor is $e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.
Re-checking the options,the correct integrating factor is $x$.

(B) The given differential equation is $\frac{dy}{dx} - 2y \tan 2x = e^x \sec 2x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -2 \tan 2x$ and $Q = e^x \sec 2x$.
First,we find the integrating factor $(IF)$:
$IF = e^{\int P dx} = e^{\int -2 \tan 2x dx} = e^{-\ln(\sec 2x)} = e^{\ln(\cos 2x)} = \cos 2x$.
The general solution is given by $y \cdot IF = \int (Q \cdot IF) dx + C$.
Substituting the values:
$y \cos 2x = \int (e^x \sec 2x \cdot \cos 2x) dx + C$.
Since $\sec 2x \cdot \cos 2x = 1$,we have:
$y \cos 2x = \int e^x dx + C$.
$y \cos 2x = e^x + C$,where $C$ is the constant of integration.

(B) The given linear differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = -\tan x$ and $Q = e^x \sec x$.
First,we find the integrating factor $(IF)$:
$IF = e^{\int P dx} = e^{\int -\tan x dx} = e^{\ln(\cos x)} = \cos x$.
The general solution is given by $y \cdot (IF) = \int Q \cdot (IF) dx + c$.
Substituting the values:
$y \cos x = \int (e^x \sec x) \cdot \cos x dx + c$.
Since $\sec x \cdot \cos x = 1$,we have:
$y \cos x = \int e^x dx + c$.
$y \cos x = e^x + c$.

(B) Given,$(x+y+1) \frac{dy}{dx} = 1$
$\Rightarrow \frac{dx}{dy} = x+y+1$
$\Rightarrow \frac{dx}{dy} - x = y+1$,which is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$.
Here,$P = -1$ and $Q = y+1$.
The Integrating Factor $(IF)$ is $e^{\int P dy} = e^{\int -1 dy} = e^{-y}$.
The solution is given by $x \cdot (IF) = \int Q \cdot (IF) dy + c$.
$x e^{-y} = \int (y+1) e^{-y} dy + c$.
Using integration by parts for $\int y e^{-y} dy$:
$x e^{-y} = [y(-e^{-y}) - \int 1 \cdot (-e^{-y}) dy] + \int e^{-y} dy + c$
$x e^{-y} = -y e^{-y} - e^{-y} - e^{-y} + c$
$x e^{-y} = -(y+2) e^{-y} + c$
Multiplying by $e^y$,we get $x = -(y+2) + ce^y$.

(A) Given the differential equation: $x^2 y - x^3 \frac{dy}{dx} = y^4 \cos x$.
Dividing both sides by $x^3 y^4$,we get:
$\frac{1}{y^3 x} - \frac{1}{y^4} \frac{dy}{dx} = \frac{\cos x}{x^3}$
Rearranging the terms:
$\frac{1}{y^4} \frac{dy}{dx} - \frac{1}{x y^3} = -\frac{\cos x}{x^3}$
Let $t = y^{-3} = \frac{1}{y^3}$. Then,$\frac{dt}{dx} = -3 y^{-4} \frac{dy}{dx}$,which implies $\frac{1}{y^4} \frac{dy}{dx} = -\frac{1}{3} \frac{dt}{dx}$.
Substituting this into the equation:
$-\frac{1}{3} \frac{dt}{dx} - \frac{t}{x} = -\frac{\cos x}{x^3}$
Multiplying by $-3$:
$\frac{dt}{dx} + \frac{3}{x} t = \frac{3 \cos x}{x^3}$
This is a linear differential equation of the form $\frac{dt}{dx} + P(x)t = Q(x)$,where $P(x) = \frac{3}{x}$ and $Q(x) = \frac{3 \cos x}{x^3}$.
The integrating factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int \frac{3}{x} dx} = e^{3 \ln x} = x^3$.
The solution is given by $t \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + c$.
$t \cdot x^3 = \int \frac{3 \cos x}{x^3} \cdot x^3 dx + c$
$t x^3 = 3 \int \cos x dx + c$
$t x^3 = 3 \sin x + c$
Since $t = y^{-3}$,we have $x^3 y^{-3} = 3 \sin x + c$.

(A) $I$. Given $dy+2xy dx=2e^{-x^2} dx$.
Dividing by $dx$,we get $\frac{dy}{dx}+2xy=2e^{-x^2}$.
This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$,where $P=2x$ and $Q=2e^{-x^2}$.
Integrating factor $I.F. = e^{\int P dx} = e^{\int 2x dx} = e^{x^2}$.
The general solution is $y(I.F.) = \int Q(I.F.) dx + c$.
$ye^{x^2} = \int 2e^{-x^2} \cdot e^{x^2} dx + c = \int 2 dx + c = 2x+c$.
Thus,statement $I$ is true.
$II$. Given $ye^{x^2}-2x=c$.
Differentiating with respect to $x$: $\frac{d}{dx}(ye^{x^2}) - \frac{d}{dx}(2x) = 0$.
$y(2x e^{x^2}) + e^{x^2} \frac{dy}{dx} - 2 = 0$.
$e^{x^2} \frac{dy}{dx} = 2 - 2xye^{x^2}$.
$\frac{dy}{dx} = 2e^{-x^2} - 2xy$.
Therefore,$dx = \frac{dy}{2e^{-x^2}-2xy}$.
Thus,statement $II$ is true.

(A) The given differential equation is $(x+2 y^3) \frac{d y}{d x}=y^2$.
Rearranging the equation to the form $\frac{d x}{d y} + P(y)x = Q(y)$,we get:
$y^2 \frac{d x}{d y} = x + 2y^3$
$\frac{d x}{d y} - \frac{1}{y^2} x = 2y$.
Here,$P(y) = -\frac{1}{y^2}$.
The integrating factor $(IF)$ is given by $e^{\int P(y) dy}$.
$IF = e^{\int -\frac{1}{y^2} dy} = e^{\int -y^{-2} dy} = e^{-(-y^{-1})} = e^{1/y}$.
Thus,the integrating factor is $e^{(1/y)}$.

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{3}$ and $Q = 1$.
The integrating factor $(IF)$ is given by $e^{\int P dx} = e^{\int \frac{1}{3} dx} = e^{x/3}$.
Multiplying both sides of the equation by the $IF$,we get:
$e^{x/3} \frac{dy}{dx} + \frac{1}{3} e^{x/3} y = e^{x/3}$
This can be written as:
$\frac{d}{dx} (y \cdot e^{x/3}) = e^{x/3}$
Integrating both sides with respect to $x$:
$y \cdot e^{x/3} = \int e^{x/3} dx + c$
$y \cdot e^{x/3} = 3e^{x/3} + c$
Dividing by $e^{x/3}$:
$y = 3 + ce^{-x/3}$

(A) We have,$\frac{dy}{dx} - y = x^2$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -1$ and $Q = x^2$.
The integrating factor is $IF = e^{\int P dx} = e^{\int -1 dx} = e^{-x}$.
The general solution is given by $y \cdot IF = \int Q \cdot IF dx + c$.
Substituting the values,we get $y e^{-x} = \int x^2 e^{-x} dx + c$.
Using integration by parts,$\int x^2 e^{-x} dx = -x^2 e^{-x} + \int 2x e^{-x} dx = -x^2 e^{-x} - 2x e^{-x} - 2e^{-x} + c$.
Thus,$y e^{-x} = -e^{-x}(x^2 + 2x + 2) + c$.
Multiplying both sides by $e^x$,we get $y = -(x^2 + 2x + 2) + ce^x$.
Rearranging the terms,we obtain $y + x^2 + 2x + 2 = ce^x$.

(C) Given the differential equation: $\cos x \frac{dy}{dx} - y \sin x = 6x$.
This can be written as the derivative of a product: $\frac{d}{dx}(y \cos x) = 6x$.
Integrating both sides with respect to $x$: $y \cos x = \int 6x \, dx = 3x^2 + C$.
Using the initial condition $y(\frac{\pi}{3}) = 0$:
$0 \cdot \cos(\frac{\pi}{3}) = 3(\frac{\pi}{3})^2 + C \Rightarrow 0 = \frac{\pi^2}{3} + C \Rightarrow C = \frac{-\pi^2}{3}$.
Thus,the general solution is $y \cos x = 3x^2 - \frac{\pi^2}{3}$.
Now,to find $y(\frac{\pi}{6})$,substitute $x = \frac{\pi}{6}$:
$y(\frac{\pi}{6}) \cos(\frac{\pi}{6}) = 3(\frac{\pi}{6})^2 - \frac{\pi^2}{3}$.
$y(\frac{\pi}{6}) \cdot \frac{\sqrt{3}}{2} = 3(\frac{\pi^2}{36}) - \frac{\pi^2}{3} = \frac{\pi^2}{12} - \frac{4\pi^2}{12} = \frac{-3\pi^2}{12} = \frac{-\pi^2}{4}$.
Therefore,$y(\frac{\pi}{6}) = \frac{-\pi^2}{4} \cdot \frac{2}{\sqrt{3}} = \frac{-\pi^2}{2 \sqrt{3}}$.

(B) Given the differential equation $\frac{dy}{dx} = \frac{1+y^2}{\tan^{-1} y - x}$.
Taking the reciprocal,we get $\frac{dx}{dy} = \frac{\tan^{-1} y - x}{1+y^2}$.
Rearranging the terms,we have $\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{\tan^{-1} y}{1+y^2}$,which is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$.
Here,$P(y) = \frac{1}{1+y^2}$ and $Q(y) = \frac{\tan^{-1} y}{1+y^2}$.
The integrating factor $IF = e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy} = e^{\tan^{-1} y}$.
The solution is given by $x \cdot IF = \int Q(y) \cdot IF dy + c$.
Substituting the values: $x e^{\tan^{-1} y} = \int \frac{\tan^{-1} y}{1+y^2} e^{\tan^{-1} y} dy + c$.
Let $t = \tan^{-1} y$,then $dt = \frac{1}{1+y^2} dy$.
The integral becomes $\int t e^t dt + c$.
Using integration by parts: $\int t e^t dt = t e^t - \int e^t dt = t e^t - e^t + c = e^t(t - 1) + c$.
Substituting $t = \tan^{-1} y$ back,we get $x e^{\tan^{-1} y} = e^{\tan^{-1} y}(\tan^{-1} y - 1) + c$.
Thus,option $B$ is correct.

(A) The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = x + y$.
Rearranging the equation,we get $\frac{dy}{dx} - y = x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -1$ and $Q = x$.
The integrating factor $(IF)$ is $e^{\int P dx} = e^{\int -1 dx} = e^{-x}$.
Multiplying both sides by the integrating factor,we get $e^{-x} \frac{dy}{dx} - e^{-x} y = x e^{-x}$.
This can be written as $\frac{d}{dx} (y e^{-x}) = x e^{-x}$.
Integrating both sides with respect to $x$,we get $y e^{-x} = \int x e^{-x} dx$.
Using integration by parts,$\int x e^{-x} dx = x(-e^{-x}) - \int 1(-e^{-x}) dx = -x e^{-x} - e^{-x} + c$.
Thus,$y e^{-x} = -x e^{-x} - e^{-x} + c$.
Multiplying by $e^x$,we get $y = -x - 1 + ce^x$,which is $y = ce^x - x - 1$.

(A) According to the question,the slope of the tangent is $\frac{dy}{dx} = x + xy$.
Rearranging the terms,we get the linear differential equation: $\frac{dy}{dx} - xy = x$.
This is of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -x$ and $Q(x) = x$.
The integrating factor $(IF)$ is $e^{\int P(x) dx} = e^{\int -x dx} = e^{-\frac{x^2}{2}}$.
The solution is given by $y \cdot IF = \int Q(x) \cdot IF dx + c$.
$y \cdot e^{-\frac{x^2}{2}} = \int x \cdot e^{-\frac{x^2}{2}} dx + c$.
Let $t = -\frac{x^2}{2}$,then $dt = -x dx$,so $x dx = -dt$.
$y \cdot e^{-\frac{x^2}{2}} = -\int e^t dt + c = -e^t + c = -e^{-\frac{x^2}{2}} + c$.
Since the curve passes through $(0, 1)$,we substitute $x=0$ and $y=1$:
$1 \cdot e^0 = -e^0 + c \Rightarrow 1 = -1 + c \Rightarrow c = 2$.
Thus,$y \cdot e^{-\frac{x^2}{2}} = -e^{-\frac{x^2}{2}} + 2$.
Dividing by $e^{-\frac{x^2}{2}}$,we get $y = -1 + 2e^{\frac{x^2}{2}}$ or $y = 2e^{\frac{x^2}{2}} - 1$.

(C) Given differential equation: $(x^2+2) dy + 2xy dx = e^x(x^2+2) dx$
Divide by $(x^2+2) dx$:
$\frac{dy}{dx} + \frac{2x}{x^2+2} y = e^x$
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{2x}{x^2+2}$ and $Q(x) = e^x$.
Integrating Factor $(IF)$ = $e^{\int P(x) dx} = e^{\int \frac{2x}{x^2+2} dx} = e^{\ln(x^2+2)} = x^2+2$.
The general solution is $y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$.
$y(x^2+2) = \int e^x(x^2+2) dx + C$.
Using integration by parts for $\int e^x(x^2+2) dx$:
$y(x^2+2) = (x^2+2)e^x - \int 2x e^x dx + C$.
$y(x^2+2) = (x^2+2)e^x - 2[x e^x - \int e^x dx] + C$.
$y(x^2+2) = (x^2+2)e^x - 2x e^x + 2e^x + C$.
$y(x^2+2) = e^x(x^2+2 - 2x + 2) + C$.
$y(x^2+2) = e^x(x^2-2x+4) + C$.

(A) Given the differential equation $\frac{d y}{d x}=\frac{2 y^2+1}{2 y^3-4 x y+y}$.
Taking the reciprocal,we get $\frac{d x}{d y}=\frac{2 y^3-4 x y+y}{2 y^2+1} = \frac{y(2 y^2+1) - 4 x y}{2 y^2+1} = y - \frac{4 x y}{2 y^2+1}$.
Rearranging the terms,we have $\frac{d x}{d y} + \left(\frac{4 y}{2 y^2+1}\right) x = y$.
This is a linear differential equation of the form $\frac{d x}{d y} + P(y) x = Q(y)$,where $P(y) = \frac{4 y}{2 y^2+1}$ and $Q(y) = y$.
The integrating factor $IF = e^{\int P(y) d y} = e^{\int \frac{4 y}{2 y^2+1} d y} = e^{\ln(2 y^2+1)} = 2 y^2+1$.
The general solution is $x(IF) = \int Q(y)(IF) d y + C$.
$x(2 y^2+1) = \int y(2 y^2+1) d y + C = \int (2 y^3+y) d y + C$.
$x(2 y^2+1) = \frac{2 y^4}{4} + \frac{y^2}{2} + C = \frac{y^4}{2} + \frac{y^2}{2} + C$.
Multiplying by $2$,we get $2 x(2 y^2+1) = y^4+y^2+2C$,which simplifies to $4 x y^2+2 x = y^4+y^2+C$.

(A) For a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,the integrating factor $(IF)$ is given by $e^{\int P(x) dx}$.
$(P)$ $(x^3+1)\frac{dy}{dx}+x^2y=3x^2 \Rightarrow \frac{dy}{dx} + \frac{x^2}{x^3+1}y = \frac{3x^2}{x^3+1}$.
$IF = e^{\int \frac{x^2}{x^3+1} dx} = e^{\frac{1}{3}\ln(x^3+1)} = (x^3+1)^{1/3}$. So,$P-5$.
$(Q)$ $x^2\frac{dy}{dx}+3xy=x^6 \Rightarrow \frac{dy}{dx} + \frac{3}{x}y = x^4$.
$IF = e^{\int \frac{3}{x} dx} = e^{3\ln x} = x^3$. So,$Q-1$.
$(R)$ $(x^3+1)^2\frac{dy}{dx}+6x^2(x^3+1)y=x^2 \Rightarrow \frac{dy}{dx} + \frac{6x^2}{x^3+1}y = \frac{x^2}{(x^3+1)^2}$.
$IF = e^{\int \frac{6x^2}{x^3+1} dx} = e^{2\ln(x^3+1)} = (x^3+1)^2$. So,$R-2$.
$(S)$ $(x^2+1)\frac{dy}{dx}+4xy=\ln x \Rightarrow \frac{dy}{dx} + \frac{4x}{x^2+1}y = \frac{\ln x}{x^2+1}$.
$IF = e^{\int \frac{4x}{x^2+1} dx} = e^{2\ln(x^2+1)} = (x^2+1)^2$. So,$S-3$.
Thus,the correct match is $P-5, Q-1, R-2, S-3$.

(D) Given the differential equation: $x y \frac{d y}{d x}=2 y^2-x^2$.
Divide by $x$: $y \frac{d y}{d x} = \frac{2 y^2}{x} - x$.
Rearranging: $y \frac{d y}{d x} - \frac{2 y^2}{x} = -x$ $\ldots$ $(i)$.
Let $v = y^2$. Then $\frac{d v}{d x} = 2 y \frac{d y}{d x}$,which implies $y \frac{d y}{d x} = \frac{1}{2} \frac{d v}{d x}$.
Substituting into $(i)$: $\frac{1}{2} \frac{d v}{d x} - \frac{2 v}{x} = -x$.
Multiply by $2$: $\frac{d v}{d x} - \frac{4 v}{x} = -2 x$.
This is a linear differential equation of the form $\frac{d v}{d x} + P(x) v = Q(x)$,where $P(x) = -\frac{4}{x}$ and $Q(x) = -2 x$.
Integrating factor $IF = e^{\int P(x) d x} = e^{\int -\frac{4}{x} d x} = e^{-4 \ln |x|} = x^{-4}$.
The solution is $v \cdot IF = \int Q(x) \cdot IF d x + c$.
$v \cdot x^{-4} = \int (-2 x) \cdot x^{-4} d x + c = \int -2 x^{-3} d x + c$.
$\frac{v}{x^4} = -2 \left( \frac{x^{-2}}{-2} \right) + c = \frac{1}{x^2} + c$.
$v = x^2 + c x^4$.
Since $v = y^2$,the family of curves is $y^2 = x^2 + c x^4$.

(A) Given the differential equation: $\cos x \frac{dy}{dx} = y \sin x - 1$.
Dividing by $\cos x$,we get: $\frac{dy}{dx} = y \tan x - \sec x$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\tan x$ and $Q(x) = -\sec x$.
The integrating factor $(IF)$ is given by $e^{\int P(x) dx} = e^{\int -\tan x dx} = e^{\ln(\cos x)} = \cos x$.
The general solution is $y \cdot IF = \int Q(x) \cdot IF dx + C$.
Substituting the values: $y \cos x = \int (-\sec x) \cdot \cos x dx + C$.
$y \cos x = \int (-1) dx + C$.
$y \cos x = -x + C$.
Given $f(0) = 1$,we substitute $x = 0$ and $y = 1$: $1 \cdot \cos(0) = -0 + C \implies 1 = C$.
Thus,$y \cos x = -x + 1$.
$y = \frac{1-x}{\cos x} = (1-x) \sec x$.
Therefore,$f(x) = (1-x) \sec x$.

(D) The given differential equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \sec x \operatorname{cosec} x = \frac{1}{\cos x \sin x} = \frac{2}{\sin 2x}$ and $Q(x) = \cos^2 x$.
First,we find the integrating factor $(IF)$:
$IF = e^{\int P(x) dx} = e^{\int \frac{1}{\sin x \cos x} dx} = e^{\int \frac{\sec^2 x}{\tan x} dx} = e^{\ln|\tan x|} = \tan x$.
The general solution is given by $y \cdot (IF) = \int Q(x) \cdot (IF) dx + c$.
$y \tan x = \int \cos^2 x \cdot \tan x dx + c$.
$y \tan x = \int \cos^2 x \cdot \frac{\sin x}{\cos x} dx + c = \int \sin x \cos x dx + c$.
$y \tan x = \frac{1}{2} \int \sin 2x dx + c = -\frac{1}{4} \cos 2x + c$.
Alternatively,using $\int \sin x \cos x dx = \frac{\sin^2 x}{2} + c$,we get $y \tan x = \frac{\sin^2 x}{2} + c$,which implies $2y \tan x = \sin^2 x + c$.

(B) The given differential equation is $(1+\cos^2 x) f'(x) - f(x) \sin 2x = 4 \sin 2x$.
Divide by $(1+\cos^2 x)$ to get the linear form $f'(x) - f(x) \frac{\sin 2x}{1+\cos^2 x} = \frac{4 \sin 2x}{1+\cos^2 x}$.
This is a linear differential equation of the form $f'(x) + P(x)f(x) = Q(x)$,where $P(x) = -\frac{\sin 2x}{1+\cos^2 x}$ and $Q(x) = \frac{4 \sin 2x}{1+\cos^2 x}$.
The integrating factor $IF = e^{\int P(x) dx} = e^{\int -\frac{\sin 2x}{1+\cos^2 x} dx}$.
Let $u = 1+\cos^2 x$,then $du = 2 \cos x (-\sin x) dx = -\sin 2x dx$.
So,$IF = e^{\int \frac{du}{u}} = e^{\ln u} = 1+\cos^2 x$.
The general solution is $f(x) \cdot IF = \int Q(x) \cdot IF dx + C$.
$f(x)(1+\cos^2 x) = \int \frac{4 \sin 2x}{1+\cos^2 x} (1+\cos^2 x) dx = \int 4 \sin 2x dx = -2 \cos 2x + C$.
Given $f(0)=0$,we have $0(1+1) = -2 \cos(0) + C \implies 0 = -2 + C \implies C = 2$.
Thus,$f(x)(1+\cos^2 x) = 2 - 2 \cos 2x = 2(1 - \cos 2x) = 4 \sin^2 x$.
$f(x) = \frac{4 \sin^2 x}{1+\cos^2 x}$.
At $x = \frac{\pi}{3}$,$\sin^2(\frac{\pi}{3}) = (\frac{\sqrt{3}}{2})^2 = \frac{3}{4}$ and $\cos^2(\frac{\pi}{3}) = (\frac{1}{2})^2 = \frac{1}{4}$.
$f(\frac{\pi}{3}) = \frac{4(3/4)}{1+1/4} = \frac{3}{5/4} = \frac{12}{5}$.

(C) Given the differential equation: $(1+y^2) dx = (\operatorname{Tan}^{-1} y - x) dy$.
Dividing both sides by $(1+y^2) dy$,we get: $\frac{dx}{dy} = \frac{\operatorname{Tan}^{-1} y - x}{1+y^2}$.
Rearranging the terms: $\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{\operatorname{Tan}^{-1} y}{1+y^2}$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{1+y^2}$ and $Q(y) = \frac{\operatorname{Tan}^{-1} y}{1+y^2}$.
The integrating factor $(IF)$ is given by $e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy} = e^{\operatorname{Tan}^{-1} y}$.
The general solution is $x \cdot (IF) = \int Q(y) \cdot (IF) dy + c$.
Substituting the values: $x \cdot e^{\operatorname{Tan}^{-1} y} = \int \frac{\operatorname{Tan}^{-1} y}{1+y^2} \cdot e^{\operatorname{Tan}^{-1} y} dy + c$.
Let $u = \operatorname{Tan}^{-1} y$,then $du = \frac{1}{1+y^2} dy$.
The integral becomes $\int u e^u du = u e^u - e^u + c$.
So,$x \cdot e^{\operatorname{Tan}^{-1} y} = \operatorname{Tan}^{-1} y \cdot e^{\operatorname{Tan}^{-1} y} - e^{\operatorname{Tan}^{-1} y} + c$.
Dividing by $e^{\operatorname{Tan}^{-1} y}$,we get $x = \operatorname{Tan}^{-1} y - 1 + c e^{-\operatorname{Tan}^{-1} y}$.
Comparing this with $x = f(y) + c e^{-\operatorname{Tan}^{-1} y}$,we find $f(y) = \operatorname{Tan}^{-1} y - 1$.

(A) Given the general solution: $y = \sin x + A \cos x$ $(i)$
Differentiating with respect to $x$: $\frac{dy}{dx} = \cos x - A \sin x$ (ii)
From $(i)$,we have $A \cos x = y - \sin x$,so $A = \frac{y - \sin x}{\cos x}$.
Substituting $A$ into (ii): $\frac{dy}{dx} = \cos x - (\frac{y - \sin x}{\cos x}) \sin x$
$\frac{dy}{dx} = \cos x - y \tan x + \sin x \tan x$
$\frac{dy}{dx} + y \tan x = \cos x + \sin x \tan x = \frac{\cos^2 x + \sin^2 x}{\cos x} = \frac{1}{\cos x} = \sec x$
Comparing this with the standard linear form $\frac{dy}{dx} + f(x)y = Q(x)$,we get $f(x) = \tan x$.
The integrating factor ($I$.$F$.) is given by $e^{\int f(x) dx} = e^{\int \tan x dx} = e^{\ln |\sec x|} = \sec x$.

(C) The given differential equation is $\frac{dy}{dx} + y f^{\prime}(x) = f(x) f^{\prime}(x)$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = f^{\prime}(x)$ and $Q(x) = f(x) f^{\prime}(x)$.
The integrating factor $IF = e^{\int P(x) dx} = e^{\int f^{\prime}(x) dx} = e^{f(x)}$.
The general solution is given by $y \cdot IF = \int Q(x) \cdot IF dx + C$.
Substituting the values,we get $y \cdot e^{f(x)} = \int f(x) f^{\prime}(x) e^{f(x)} dx + C$.
Let $u = f(x)$,then $du = f^{\prime}(x) dx$. The integral becomes $\int u e^u du$.
Using integration by parts,$\int u e^u du = u e^u - e^u$.
So,$y \cdot e^{f(x)} = f(x) e^{f(x)} - e^{f(x)} + C$.
Dividing by $e^{f(x)}$,we get $y = f(x) - 1 + C e^{-f(x)}$.