Linear differential equations Questions in English

(B) Given the differential equation: $(x + 2y^3) \frac{dy}{dx} = y$.
Rearranging the terms,we get: $\frac{dx}{dy} = \frac{x + 2y^3}{y}$.
This can be written as: $\frac{dx}{dy} - \frac{1}{y}x = 2y^2$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = 2y^2$.
The Integrating Factor $(I.F.)$ is given by: $I.F. = e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln y} = \frac{1}{y}$.
The general solution is: $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$.
Substituting the values: $x \cdot \frac{1}{y} = \int 2y^2 \cdot \frac{1}{y} dy + c$.
$\frac{x}{y} = \int 2y dy + c$.
$\frac{x}{y} = y^2 + c$.

(A) Given the differential equation: $x^2 \frac{dy}{dx} \cos \frac{1}{x} - y \sin \frac{1}{x} = -1$.
Dividing by $x^2 \cos \frac{1}{x}$,we get: $\frac{dy}{dx} - y \frac{\tan(1/x)}{x^2} = -\frac{\sec(1/x)}{x^2}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\frac{\tan(1/x)}{x^2}$ and $Q = -\frac{\sec(1/x)}{x^2}$.
The Integrating Factor $(IF)$ is $e^{\int P dx} = e^{\int -\frac{\tan(1/x)}{x^2} dx}$. Let $u = \frac{1}{x}$,then $du = -\frac{1}{x^2} dx$.
$IF = e^{\int \tan u du} = e^{\ln |\sec u|} = \sec(\frac{1}{x})$.
The solution is $y \cdot IF = \int Q \cdot IF dx + c$.
$y \sec(\frac{1}{x}) = \int -\frac{\sec(1/x)}{x^2} \cdot \sec(\frac{1}{x}) dx = -\int \sec^2(\frac{1}{x}) \cdot \frac{1}{x^2} dx$.
Let $u = \frac{1}{x}$,then $du = -\frac{1}{x^2} dx$.
$y \sec(\frac{1}{x}) = \int \sec^2 u du = \tan u + c = \tan(\frac{1}{x}) + c$.
As $x \rightarrow \infty$,$\frac{1}{x} \rightarrow 0$. Given $y \rightarrow -1$,we have: $-1 \cdot \sec(0) = \tan(0) + c \Rightarrow -1 = 0 + c \Rightarrow c = -1$.
Thus,$y \sec(\frac{1}{x}) = \tan(\frac{1}{x}) - 1$.
$y = \frac{\tan(1/x) - 1}{\sec(1/x)} = \sin(\frac{1}{x}) - \cos(\frac{1}{x})$.

(B) The given linear differential equation is $(x + 1)f'(x) - 2x(x + 1)f(x) = \frac{e^{x^2}}{x + 1}$.
Dividing by $(x + 1)$,we get $f'(x) - 2xf(x) = \frac{e^{x^2}}{(x + 1)^2}$.
The integrating factor $I.F. = e^{\int -2x \, dx} = e^{-x^2}$.
Multiplying both sides by $I.F.$,we have $\frac{d}{dx} [f(x) e^{-x^2}] = \frac{e^{x^2}}{(x + 1)^2} e^{-x^2} = \frac{1}{(x + 1)^2}$.
Integrating both sides,$f(x) e^{-x^2} = \int \frac{1}{(x + 1)^2} \, dx = -\frac{1}{x + 1} + C$.
Given $f(0) = 5$,we have $5(e^0) = -\frac{1}{0 + 1} + C \Rightarrow 5 = -1 + C \Rightarrow C = 6$.
Thus,$f(x) e^{-x^2} = 6 - \frac{1}{x + 1} = \frac{6x + 6 - 1}{x + 1} = \frac{6x + 5}{x + 1}$.
Therefore,$f(x) = \left( \frac{6x + 5}{x + 1} \right) e^{x^2}$.

(A) Given the equation $\int\limits_a^x {t\,y(t)dt} = x^2 + y(x)$.
Differentiating both sides with respect to $x$ using the Leibniz integral rule:
$x\,y(x) = 2x + y'(x)$
Rearranging the terms to form a linear differential equation:
$y'(x) + x\,y(x) = 2x$
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = x$ and $Q(x) = 2x$.
The integrating factor $(I.F.)$ is:
$I.F. = e^{\int x\,dx} = e^{\frac{x^2}{2}}$
Multiplying both sides by the $I.F.$:
$e^{\frac{x^2}{2}} \frac{dy}{dx} + x\,e^{\frac{x^2}{2}} y = 2x\,e^{\frac{x^2}{2}}$
$\frac{d}{dx} \left( y\,e^{\frac{x^2}{2}} \right) = 2x\,e^{\frac{x^2}{2}}$
Integrating both sides with respect to $x$:
$y\,e^{\frac{x^2}{2}} = \int 2x\,e^{\frac{x^2}{2}} dx + C$
Let $u = \frac{x^2}{2}$,then $du = x\,dx$:
$y\,e^{\frac{x^2}{2}} = 2\,e^{\frac{x^2}{2}} + C$
Dividing by $e^{\frac{x^2}{2}}$:
$y = 2 + C\,e^{-\frac{x^2}{2}}$
To find $C$,use the initial condition at $x = a$:
$\int\limits_a^a {t\,y(t)dt} = a^2 + y(a) \implies 0 = a^2 + y(a) \implies y(a) = -a^2$
Substituting $x = a$ into the general solution:
$-a^2 = 2 + C\,e^{-\frac{a^2}{2}}$
$C\,e^{-\frac{a^2}{2}} = -(2 + a^2)$
$C = -(2 + a^2)e^{\frac{a^2}{2}}$
Substituting $C$ back into the general solution:
$y = 2 - (2 + a^2)e^{\frac{a^2}{2}} e^{-\frac{x^2}{2}}$
$y = 2 - (2 + a^2)e^{\frac{a^2 - x^2}{2}}$
Note: The provided options use $e^{\frac{x^2 - a^2}{2}}$. Re-evaluating the sign of the derivative: $x\,y(x) = 2x + y'(x) \implies y' - x\,y = -2x$. $I.F. = e^{-\frac{x^2}{2}}$. $y\,e^{-\frac{x^2}{2}} = \int -2x\,e^{-\frac{x^2}{2}} dx = 2\,e^{-\frac{x^2}{2}} + C$. $y = 2 + C\,e^{\frac{x^2}{2}}$. At $x=a, y=-a^2 \implies -a^2 = 2 + C\,e^{\frac{a^2}{2}} \implies C = -(2+a^2)e^{-\frac{a^2}{2}}$. Thus $y = 2 - (2+a^2)e^{\frac{x^2-a^2}{2}}$.

(A) Let $y^2 = t$. Then $2y \frac{{dy}}{{dx}} = \frac{{dt}}{{dx}}$.
Substituting this into the differential equation:
$\left( {{e^{{x^2}}} + {e^t}} \right) \frac{1}{2} \frac{{dt}}{{dx}} + {e^{{x^2}}} x (t - 1) = 0$
$\left( {{e^{{x^2}}} + {e^t}} \right) \frac{{dt}}{{dx}} + 2 x {e^{{x^2}}} (t - 1) = 0$
Rearranging the terms:
$\frac{{dx}}{{dt}} (e^t + {e^{{x^2}}}) + 2 x {e^{{x^2}}} (t - 1) = 0$
Let $z = {e^{{x^2}}}$,then $\frac{{dz}}{{dt}} = {e^{{x^2}}} \cdot 2x \frac{{dx}}{{dt}}$.
Substituting $z$ into the equation:
$\frac{{dz}}{{dt}} + z + {e^t} + z(t - 1) \frac{{dx}}{{dt}} \cdot \frac{1}{x} \dots$ (This approach is complex,let's use the substitution $z = {e^{{x^2}}}$ directly).
Given $\left( {{e^{{x^2}}} + {e^{{y^2}}}} \right) y \frac{{dy}}{{dx}} + {e^{{x^2}}} x (y^2 - 1) = 0$.
Let $y^2 = t$,then $2y \frac{{dy}}{{dx}} = \frac{{dt}}{{dx}}$.
$\frac{1}{2} (e^{{x^2}} + e^t) \frac{{dt}}{{dx}} + x e^{{x^2}} (t - 1) = 0$.
Let $u = e^{{x^2}}$,then $\frac{{du}}{{dx}} = 2x e^{{x^2}} \implies du = 2x e^{{x^2}} dx$.
This is a linear differential equation in $z = e^{{x^2}}$ with respect to $t$:
$\frac{{dz}}{{dt}} + \frac{z}{t-1} = -\frac{e^t}{t-1}$.
Integrating factor $I.F. = e^{\int \frac{1}{t-1} dt} = t-1$.
$z(t-1) = \int -e^t dt = -e^t + C$.
Substituting back $z = e^{{x^2}}$ and $t = y^2$:
$e^{{x^2}}(y^2 - 1) + e^{{y^2}} = C$.

(D) differential equation is linear if the dependent variable and its derivatives appear only in the first degree and are not multiplied together.
$1$. For option $(A)$: $\frac{dy}{dx} + \frac{y}{x} = \ln x$. This is a first-order linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = \ln x$. Thus,it is linear.
$2$. For option $(B)$: $\frac{d^2y}{dx^2} = \cos x$. This is a second-order linear differential equation because the dependent variable $y$ and its derivative $\frac{d^2y}{dx^2}$ are of the first degree. Thus,it is linear.
$3$. For option $(C)$: $dx + dy = 0$. This can be rewritten as $1 + \frac{dy}{dx} = 0$ or $\frac{dy}{dx} = -1$. This is a first-order linear differential equation. Thus,it is linear.
Since all three equations satisfy the definition of a linear differential equation,the correct answer is $(D)$.

(D) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \cos x$ and $Q = \cos x$.
The Integrating Factor $(I.F.)$ is given by $e^{\int P \, dx} = e^{\int \cos x \, dx} = e^{\sin x}$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + C$.
Substituting the values: $y \cdot e^{\sin x} = \int \cos x \cdot e^{\sin x} \, dx + C$.
Let $u = \sin x$,then $du = \cos x \, dx$. The integral becomes $\int e^u \, du = e^u = e^{\sin x}$.
So,$y \cdot e^{\sin x} = e^{\sin x} + C$.
Dividing by $e^{\sin x}$,we get $y = 1 + C e^{-\sin x}$.
Since the curve passes through $(0, 1)$,we substitute $x = 0$ and $y = 1$: $1 = 1 + C e^{-\sin 0} \implies 1 = 1 + C(1) \implies C = 0$.
Thus,the function is $y = 1$.
Since $y = 1$ is a constant function,it is also periodic (with any period $T > 0$) and it is continuous and differentiable for all $x \in \mathbb{R}$.
Therefore,all the given statements are correct.

(D) The given differential equation is $\frac{dy}{dx} \sin x - y \cos x = -\frac{\sin^2 x}{x^2}$.
Dividing by $\sin^2 x$,we get $\frac{1}{\sin x} \frac{dy}{dx} - y \frac{\cos x}{\sin^2 x} = -\frac{1}{x^2}$.
This is equivalent to $\frac{d}{dx} \left( \frac{y}{\sin x} \right) = -\frac{1}{x^2}$.
Integrating both sides,we get $\frac{y}{\sin x} = \int -\frac{1}{x^2} dx = \frac{1}{x} + C$.
So,$y = \sin x \left( \frac{1}{x} + C \right)$.
Given $y \rightarrow 0$ as $x \rightarrow \infty$,we must have $C = 0$,so $f(x) = \frac{\sin x}{x}$.
$1$. $\mathop {Lim}\limits_{x \to 0} f(x) = \mathop {Lim}\limits_{x \to 0} \frac{\sin x}{x} = 1$.
$2$. Since $\sin x < x$ for $x > 0$,$f(x) = \frac{\sin x}{x} < 1$. Thus,$\int_0^{\pi/2} f(x) dx < \int_0^{\pi/2} 1 dx = \frac{\pi}{2}$.
$3$. Using the series expansion $\frac{\sin x}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \dots$,the integral $\int_0^{\pi/2} (1 - \frac{x^2}{6}) dx = [x - \frac{x^3}{18}]_0^{\pi/2} = \frac{\pi}{2} - \frac{\pi^3}{144} \approx 1.57 - 0.21 = 1.36 > 1$.
Thus,all statements are correct.

(A) Given $y = (A + Bx) e^{mx} + (m - 1)^{-2} e^x$.
Let $y = y_1 + y_2$,where $y_1 = (A + Bx) e^{mx}$ and $y_2 = (m - 1)^{-2} e^x$.
For $y_1 = (A + Bx) e^{mx}$,the differential equation satisfied is $\frac{d^2y_1}{dx^2} - 2m \frac{dy_1}{dx} + m^2y_1 = 0$.
Now,consider $y_2 = (m - 1)^{-2} e^x$.
$\frac{dy_2}{dx} = (m - 1)^{-2} e^x$ and $\frac{d^2y_2}{dx^2} = (m - 1)^{-2} e^x$.
Substituting $y_2$ into the expression $\frac{d^2y_2}{dx^2} - 2m \frac{dy_2}{dx} + m^2y_2$:
$= (m - 1)^{-2} e^x - 2m(m - 1)^{-2} e^x + m^2(m - 1)^{-2} e^x$
$= (m - 1)^{-2} e^x [1 - 2m + m^2]$
$= (m - 1)^{-2} e^x (m - 1)^2$
$= e^x$.
Thus,$\frac{d^2y}{dx^2} - 2m \frac{dy}{dx} + m^2y = 0 + e^x = e^x$.

(A) Given the equation: $f(x) = f'(x) + f''(x) + f'''(x) + \dots \infty$
Taking the derivative of both sides with respect to $x$:
$f'(x) = f''(x) + f'''(x) + f''''(x) + \dots \infty$
Adding $f'(x)$ to both sides of the original equation:
$f(x) + f'(x) = f'(x) + f''(x) + f'''(x) + \dots + f'(x)$
This simplifies to:
$f(x) + f'(x) = 2f'(x)$
$f(x) = f'(x)$
This is a first-order linear differential equation:
$\frac{f'(x)}{f(x)} = 1$
Integrating both sides with respect to $x$:
$\ln|f(x)| = x + C$
Given $f(0) = 1$:
$\ln(1) = 0 + C \Rightarrow C = 0$
Thus,$\ln|f(x)| = x$,which implies $f(x) = e^x$.
Wait,re-evaluating the series:
$f(x) = f'(x) + f''(x) + f'''(x) + \dots$
$f'(x) = f''(x) + f'''(x) + f''''(x) + \dots$
Subtracting the second from the first:
$f(x) - f'(x) = f'(x)$
$f(x) = 2f'(x)$
$\frac{f'(x)}{f(x)} = \frac{1}{2}$
Integrating: $\ln f(x) = \frac{x}{2} + C$
Since $f(0) = 1$,$C = 0$.
Therefore,$f(x) = e^{x/2}$.

(B) Given the differential equation: $\frac{dy}{dx} + x \sin^2 y = \sin y \cos y$
Divide both sides by $\sin^2 y$: $\csc^2 y \frac{dy}{dx} + x = \cot y$
Rearrange the terms: $\csc^2 y \frac{dy}{dx} - \cot y = -x$
Let $v = \cot y$. Then $\frac{dv}{dx} = -\csc^2 y \frac{dy}{dx}$,which implies $-\frac{dv}{dx} = \csc^2 y \frac{dy}{dx}$.
Substituting this into the equation: $-\frac{dv}{dx} - v = -x$,or $\frac{dv}{dx} + v = x$.
This is a linear differential equation of the form $\frac{dv}{dx} + Pv = Q$,where $P = 1$ and $Q = x$.
The integrating factor is $IF = e^{\int 1 dx} = e^x$.
The general solution is $v \cdot e^x = \int x e^x dx + C$.
Using integration by parts: $\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x = (x - 1)e^x$.
So,$v e^x = (x - 1)e^x + C$.
Dividing by $e^x$: $v = (x - 1) + Ce^{-x}$.
Substituting $v = \cot y$ back: $\cot y = (x - 1) + Ce^{-x}$.

(D) The given differential equation is $y \sin 2x - \cos x + (1 + \sin^2 x) \frac{dy}{dx} = 0$,where $y = f(x)$.
Rearranging the terms,we get $\frac{dy}{dx} + \left(\frac{\sin 2x}{1 + \sin^2 x}\right) y = \frac{\cos x}{1 + \sin^2 x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{\sin 2x}{1 + \sin^2 x}$ and $Q(x) = \frac{\cos x}{1 + \sin^2 x}$.
The Integrating Factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int \frac{\sin 2x}{1 + \sin^2 x} dx}$.
Let $u = 1 + \sin^2 x$,then $du = 2 \sin x \cos x dx = \sin 2x dx$. Thus,$\int P(x) dx = \int \frac{du}{u} = \ln(1 + \sin^2 x)$.
So,$I$.$F$. $= e^{\ln(1 + \sin^2 x)} = 1 + \sin^2 x$.
The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.
$y(1 + \sin^2 x) = \int \frac{\cos x}{1 + \sin^2 x} \cdot (1 + \sin^2 x) dx + C = \int \cos x dx + C = \sin x + C$.
Given $f(0) = 0$,we have $0(1 + 0) = \sin(0) + C$,which implies $C = 0$.
Thus,$f(x) = \frac{\sin x}{1 + \sin^2 x}$.
For $x = \frac{\pi}{6}$,$f(\frac{\pi}{6}) = \frac{\sin(\pi/6)}{1 + \sin^2(\pi/6)} = \frac{1/2}{1 + (1/2)^2} = \frac{1/2}{1 + 1/4} = \frac{1/2}{5/4} = \frac{1}{2} \cdot \frac{4}{5} = \frac{2}{5}$.

(A) Given the equation: $\int_{a}^{x} t y(t) dt = x^2 + y(x)$ $(1)$
Differentiating both sides with respect to $x$ using Leibniz's rule:
$x y(x) = 2x + \frac{dy}{dx}$
Rearranging the terms:
$\frac{dy}{dx} = x y - 2x = x(y - 2)$
Separating the variables:
$\int \frac{dy}{y - 2} = \int x dx$
Integrating both sides:
$\ln |y - 2| = \frac{x^2}{2} + C$
$y - 2 = K e^{\frac{x^2}{2}}$ (where $K = \pm e^C$)
At $x = a$,the integral $\int_{a}^{a} t y(t) dt = 0$,so from $(1)$:
$0 = a^2 + y(a) \Rightarrow y(a) = -a^2$
Substituting $x = a$ and $y = -a^2$ into the general solution:
$-a^2 - 2 = K e^{\frac{a^2}{2}}$
$K = -(a^2 + 2) e^{-\frac{a^2}{2}}$
Substituting $K$ back into the equation:
$y - 2 = -(a^2 + 2) e^{-\frac{a^2}{2}} \cdot e^{\frac{x^2}{2}}$
$y - 2 = -(a^2 + 2) e^{\frac{x^2 - a^2}{2}}$
$y = 2 - (2 + a^2) e^{\frac{x^2 - a^2}{2}}$

(C) Given the differential equation: $(x + 2y^3)\frac{dy}{dx} - y = 0$
Rearranging the terms,we get: $(x + 2y^3)dy = y dx$
$\Rightarrow \frac{dx}{dy} = \frac{x + 2y^3}{y}$
$\Rightarrow \frac{dx}{dy} = \frac{x}{y} + 2y^2$
$\Rightarrow \frac{dx}{dy} - \frac{1}{y}x = 2y^2$
This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = -\frac{1}{y}$ and $Q = 2y^2$.
The Integrating Factor ($I$.$F$.) is given by: $I.F. = e^{\int P dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln y} = \frac{1}{y}$.
The general solution is given by: $x \cdot (I.F.) = \int Q \cdot (I.F.) dy + C$
$\Rightarrow x \cdot \frac{1}{y} = \int 2y^2 \cdot \frac{1}{y} dy + C$
$\Rightarrow \frac{x}{y} = \int 2y dy + C$
$\Rightarrow \frac{x}{y} = y^2 + C$
$\Rightarrow x = y^3 + Cy$
Thus,the solution is $x = y^3 + Cy$.

(A) Given the differential equation $\frac{dx}{dy} = \frac{x}{1 + x e^y \cos(x^2)}$.
Taking the reciprocal,we get $\frac{dy}{dx} = \frac{1 + x e^y \cos(x^2)}{x} = \frac{1}{x} + e^y \cos(x^2)$.
Rearranging the terms,we have $\frac{dy}{dx} - e^y \cos(x^2) = \frac{1}{x}$.
This is not a standard linear form. Let's re-examine the original equation: $\frac{dx}{dy} = \frac{x}{1 + x e^y \cos(x^2)}$.
$\frac{dy}{dx} = \frac{1 + x e^y \cos(x^2)}{x} = \frac{1}{x} + e^y \cos(x^2)$.
Divide by $e^y$: $e^{-y} \frac{dy}{dx} - \cos(x^2) = \frac{1}{x} e^{-y}$.
$e^{-y} \frac{dy}{dx} - \frac{e^{-y}}{x} = \cos(x^2)$.
Let $v = -e^{-y}$,then $\frac{dv}{dx} = e^{-y} \frac{dy}{dx}$.
Substituting this into the equation: $\frac{dv}{dx} + \frac{v}{x} = \cos(x^2)$.
This is a linear differential equation in $v$ with integrating factor $I.F. = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.
The solution is $v \cdot x = \int x \cos(x^2) dx$.
Let $u = x^2$,then $du = 2x dx$,so $x dx = \frac{1}{2} du$.
$v \cdot x = \int \frac{1}{2} \cos(u) du = \frac{1}{2} \sin(u) + c = \frac{1}{2} \sin(x^2) + c$.
Substituting $v = -e^{-y}$: $-x e^{-y} = \frac{1}{2} \sin(x^2) + c$.
$-2x = e^y (\sin(x^2) + 2c)$.
Rearranging gives $2x + e^y(C + \sin(x^2)) = 0$.

(C) Given the equation: $\frac{f(x)}{1+x^2} = 1 + \int_{0}^{x} \frac{f(t)}{1+t^2} dt$.
Let $g(x) = \int_{0}^{x} \frac{f(t)}{1+t^2} dt$. Then $g'(x) = \frac{f(x)}{1+x^2}$.
The equation becomes $\frac{f(x)}{1+x^2} = 1 + g(x)$.
Substituting $f(x) = (1+x^2)g'(x)$ into the equation: $g'(x) = 1 + g(x)$.
This is a linear differential equation: $\frac{dg}{dx} - g = 1$.
The integrating factor is $e^{\int -1 dx} = e^{-x}$.
Multiplying by $e^{-x}$: $\frac{d}{dx}(g(x)e^{-x}) = e^{-x}$.
Integrating both sides: $g(x)e^{-x} = -e^{-x} + C$.
So,$g(x) = C e^x - 1$.
Since $g(0) = \int_{0}^{0} \dots = 0$,we have $0 = C - 1$,so $C = 1$.
Thus,$g(x) = e^x - 1$.
Then $g'(x) = e^x$.
Since $f(x) = (1+x^2)g'(x)$,we have $f(x) = (1+x^2)e^x$.
Therefore,$f(1) = (1+1^2)e^1 = 2e$.

(B) Given the differential equation: $\frac{dy}{dx} + \frac{1}{x} \sin 2y = x^3 \cos^2 y$.
Using $\sin 2y = 2 \sin y \cos y$,we have $\frac{dy}{dx} + \frac{2}{x} \sin y \cos y = x^3 \cos^2 y$.
Dividing both sides by $\cos^2 y$,we get: $\sec^2 y \frac{dy}{dx} + \frac{2}{x} \tan y = x^3$.
Let $z = \tan y$,then $\frac{dz}{dx} = \sec^2 y \frac{dy}{dx}$.
The equation becomes: $\frac{dz}{dx} + \frac{2}{x} z = x^3$.
This is a linear differential equation of the form $\frac{dz}{dx} + P(x)z = Q(x)$,where $P(x) = \frac{2}{x}$ and $Q(x) = x^3$.
The integrating factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln x} = x^2$.
The solution is $z \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.
$z \cdot x^2 = \int x^3 \cdot x^2 dx + C = \int x^5 dx + C$.
$z x^2 = \frac{x^6}{6} + C$.
Substituting $z = \tan y$: $(\tan y) x^2 = \frac{x^6}{6} + C$.
Multiplying by $6$: $6x^2 \tan y = x^6 + 6C$.
Since $6C$ is an arbitrary constant,we can write it as $C$.
Thus,$6x^2 \tan y = x^6 + C$.

(D) The given differential equation is a linear differential equation of the form $y' + Py = Q$,where $P = 1$ and $Q = 2(\sin x + \cos x)$.
The integrating factor $(IF)$ is $e^{\int P dx} = e^{\int 1 dx} = e^x$.
The general solution is given by $y \cdot IF = \int Q \cdot IF dx + C$.
$y \cdot e^x = \int 2(\sin x + \cos x) e^x dx + C$.
Using the identity $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$,where $f(x) = \sin x$ and $f'(x) = \cos x$,we get:
$y \cdot e^x = 2 e^x \sin x + C$.
Dividing by $e^x$,we get $y = 2 \sin x + C e^{-x}$.
Given $y(0) = 1$,we substitute $x = 0$ and $y = 1$:
$1 = 2 \sin(0) + C e^0 \Rightarrow 1 = 0 + C \Rightarrow C = 1$.
Thus,the particular solution is $y = 2 \sin x + e^{-x}$.
Now,evaluating at $x = \pi$:
$y(\pi) = 2 \sin(\pi) + e^{-\pi} = 2(0) + e^{-\pi} = e^{-\pi}$.

(A) Given the differential equation: $\frac{dy}{dx} = \frac{1}{xy(x^2 \sin y^2 + 1)}$.
Rearranging,we get: $\frac{dx}{dy} = xy(x^2 \sin y^2 + 1) = x^3 y \sin y^2 + xy$.
Divide by $x^3$: $x^{-3} \frac{dx}{dy} - yx^{-1} = y \sin y^2$.
Let $u = x^{-2}$,then $\frac{du}{dy} = -2x^{-3} \frac{dx}{dy}$,so $x^{-3} \frac{dx}{dy} = -\frac{1}{2} \frac{du}{dy}$.
Substituting into the equation: $-\frac{1}{2} \frac{du}{dy} - yu = y \sin y^2 \Rightarrow \frac{du}{dy} + 2yu = -2y \sin y^2$.
This is a linear differential equation in $u$ with integrating factor $IF = e^{\int 2y dy} = e^{y^2}$.
The solution is $u e^{y^2} = \int (-2y \sin y^2) e^{y^2} dy$.
Let $t = y^2$,then $dt = 2y dy$. The integral becomes $\int -\sin t e^t dt$.
Using integration by parts: $\int e^t \sin t dt = \frac{e^t}{2} (\sin t - \cos t)$.
So,$u e^{y^2} = -\frac{e^{y^2}}{2} (\sin y^2 - \cos y^2) + C$.
Substituting $u = x^{-2}$: $\frac{e^{y^2}}{x^2} = \frac{e^{y^2}}{2} (\cos y^2 - \sin y^2) + C$.
Rearranging gives: $e^{y^2} \left( \frac{1}{x^2} - \frac{\cos y^2}{2} + \frac{\sin y^2}{2} \right) = C$.

(D) Given the inequality $f(x) + f'(x) \le 1$.
Multiply both sides by the integrating factor $e^x$:
$e^x f'(x) + e^x f(x) \le e^x$
This can be written as the derivative of a product:
$\frac{d}{dx} (f(x) e^x) \le e^x$
Now,integrate both sides from $0$ to $1$:
$\int_{0}^{1} \frac{d}{dx} (f(x) e^x) dx \le \int_{0}^{1} e^x dx$
$[f(x) e^x]_{0}^{1} \le [e^x]_{0}^{1}$
$f(1) e^1 - f(0) e^0 \le e^1 - e^0$
Since $f(0) = 0$,we have:
$f(1) e \le e - 1$
$f(1) \le \frac{e-1}{e}$
Thus,the largest possible value of $f(1)$ is $\frac{e-1}{e}$.

(A) Given the differential equation: $ydx + y^2dy = xdy$.
Rearranging the terms to form a linear differential equation in $x$:
$ydx = (x - y^2)dy$
$\frac{dx}{dy} = \frac{x - y^2}{y} = \frac{x}{y} - y$
$\frac{dx}{dy} - \frac{1}{y}x = -y$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = -y$.
The integrating factor $(IF)$ is given by $IF = e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = \frac{1}{y}$.
The general solution is $x \cdot IF = \int Q(y) \cdot IF dy + C$.
$x \cdot \frac{1}{y} = \int (-y) \cdot \frac{1}{y} dy + C$
$\frac{x}{y} = \int -1 dy + C = -y + C$.
Using the initial condition $y(1) = 1$,we substitute $x = 1$ and $y = 1$:
$\frac{1}{1} = -1 + C \Rightarrow C = 2$.
So,the equation is $\frac{x}{y} = -y + 2$,which implies $x = 2y - y^2$.
To find $y(-3)$,we set $x = -3$:
$-3 = 2y - y^2 \Rightarrow y^2 - 2y - 3 = 0$.
$(y - 3)(y + 1) = 0$.
Since $y > 0$,we have $y = 3$.

(C) The given differential equation is $(x^3 + 1)dy = x(1 - 3xy)dx$.
Rearranging the terms,we get $(x^3 + 1)dy + 3x^2ydx = xdx$.
This can be written as $d(y(x^3 + 1)) = xdx$.
Integrating both sides,we get $y(x^3 + 1) = \int xdx = \frac{x^2}{2} + C$.
Given $f(0) = 0$,substituting $x = 0$ and $y = 0$ gives $0(0 + 1) = 0 + C$,so $C = 0$.
Thus,$y(x^3 + 1) = \frac{x^2}{2}$,which implies $f(x) = \frac{x^2}{2(x^3 + 1)}$.
Now,we calculate the limit: $\mathop {\lim }\limits_{x \to 0} \frac{x^2}{f(x)} = \mathop {\lim }\limits_{x \to 0} \frac{x^2}{\frac{x^2}{2(x^3 + 1)}} = \mathop {\lim }\limits_{x \to 0} 2(x^3 + 1) = 2(0 + 1) = 2$.

(C) Let the point of tangency be $(x, y)$. The equation of the tangent line is $Y - y = \frac{dy}{dx}(X - x)$.
To find the $y$-intercept,set $X = 0$:
$Y - y = \frac{dy}{dx}(0 - x) \implies Y = y - x \frac{dy}{dx}$.
According to the problem,the $y$-intercept $Y$ is proportional to the square of the ordinate $y$,so $Y = ky^2$ for some constant $k$.
Equating the two expressions for $Y$:
$y - x \frac{dy}{dx} = ky^2$.
Rearranging the terms:
$x \frac{dy}{dx} - y = -ky^2$.
Divide by $xy^2$:
$\frac{1}{y^2} \frac{dy}{dx} - \frac{1}{xy} = -\frac{k}{x}$.
Let $v = \frac{1}{y}$,then $\frac{dv}{dx} = -\frac{1}{y^2} \frac{dy}{dx}$.
Substituting this into the equation:
$-\frac{dv}{dx} - \frac{v}{x} = -\frac{k}{x} \implies \frac{dv}{dx} + \frac{v}{x} = \frac{k}{x}$.
This is a linear differential equation. The integrating factor is $e^{\int \frac{1}{x} dx} = x$.
Multiplying by $x$:
$x \frac{dv}{dx} + v = k \implies \frac{d}{dx}(xv) = k$.
Integrating both sides:
$xv = kx + C \implies x(\frac{1}{y}) = kx + C$.
Dividing by $x$:
$\frac{1}{y} = k + \frac{C}{x} \implies \frac{C}{x} + \frac{1}{y} = k$.
Rearranging into the form $\frac{c_1}{x} + \frac{c_2}{y} = 1$ by dividing by $k$ (where $c_1 = C/k$ and $c_2 = 1/k$).

(B) Given the differential equation $\frac{dy}{dx} = \frac{y}{y^2 - x}$.
Taking the reciprocal,we get $\frac{dx}{dy} = \frac{y^2 - x}{y} = y - \frac{x}{y}$.
Rearranging the terms,we obtain the linear differential equation in $x$: $\frac{dx}{dy} + \frac{1}{y}x = y$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{y}$ and $Q(y) = y$.
The integrating factor $(IF)$ is given by $IF = e^{\int P(y) dy} = e^{\int \frac{1}{y} dy} = e^{\ln|y|} = y$.
The general solution is $x \cdot (IF) = \int Q(y) \cdot (IF) dy + c$.
Substituting the values,$x \cdot y = \int y \cdot y dy + c$.
$xy = \int y^2 dy + c$.
$xy = \frac{y^3}{3} + c$.
Multiplying by $3$,we get $3xy = y^3 + 3c$,which can be written as $y^3 - 3xy = c'$ (where $c' = -3c$).
Thus,the correct option is $B$.

(C) Given that $\lim_{x \to \infty} (f(x) + f'(x) + f''(x)) = 5$.
Let $L = \lim_{x \to \infty} f(x)$. Since $f(x)$ is bounded in $(2, 20)$,the limit exists.
Consider the differential equation $f''(x) + f'(x) + f(x) = g(x)$.
As $x \to \infty$,$g(x) \to 5$.
For the linear differential equation with constant coefficients,the particular solution for the constant term $5$ is $f(x) = c$.
Substituting $f(x) = c$ into the equation,we get $0 + 0 + c = 5$,so $c = 5$.
Thus,$\lim_{x \to \infty} f(x) = 5$.
Since $\lim_{x \to \infty} g(x) = 5$,we have $\lim_{x \to \infty} (f(x) - g(x)) = 5 - 5 = 0$.

(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{2}{x}$ and $Q(x) = x^3$.
First,we find the integrating factor $(IF)$:
$IF = e^{\int P(x) dx} = e^{\int -\frac{2}{x} dx} = e^{-2 \ln|x|} = e^{\ln|x^{-2}|} = x^{-2} = \frac{1}{x^2}$.
The general solution is given by $y \times (IF) = \int Q(x) \times (IF) dx + c$.
Substituting the values:
$y \times \frac{1}{x^2} = \int x^3 \times \frac{1}{x^2} dx + c$
$y \times \frac{1}{x^2} = \int x dx + c$
$y \times \frac{1}{x^2} = \frac{x^2}{2} + c$
Multiplying both sides by $x^2$:
$y = \frac{x^4}{2} + cx^2$
$2y = x^4 + 2cx^2$.
Since $2c$ is an arbitrary constant,we can write it as $C$.
Thus,$2y = x^4 + Cx^2$.

(B) The given differential equation is $x \frac{dy}{dx} - 2y = x^2 + \sin \left( \frac{1}{x^2} \right)$.
Divide the entire equation by $x$ to write it in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$:
$\frac{dy}{dx} - \frac{2}{x}y = x + \frac{1}{x} \sin \left( \frac{1}{x^2} \right)$.
Here,$P(x) = -\frac{2}{x}$.
The integrating factor $(IF)$ is given by the formula $IF = e^{\int P(x) dx}$.
$IF = e^{\int -\frac{2}{x} dx} = e^{-2 \ln|x|} = e^{\ln|x^{-2}|} = x^{-2} = \frac{1}{x^2}$.

(C) Given the differential equation: $\frac{dy}{dx} + \frac{y \ln y}{x} = \frac{y(\ln y)^2}{x^2}$.
Divide both sides by $y(\ln y)^2$: $\frac{1}{y(\ln y)^2} \frac{dy}{dx} + \frac{1}{x \ln y} = \frac{1}{x^2}$.
Let $t = \frac{1}{\ln y}$. Then $\frac{dt}{dx} = -\frac{1}{y(\ln y)^2} \frac{dy}{dx}$.
Substituting this into the equation,we get: $-\frac{dt}{dx} + \frac{t}{x} = \frac{1}{x^2}$,which simplifies to $\frac{dt}{dx} - \frac{t}{x} = -\frac{1}{x^2}$.
This is a linear differential equation of the form $\frac{dt}{dx} + P(x)t = Q(x)$,where $P(x) = -\frac{1}{x}$ and $Q(x) = -\frac{1}{x^2}$.
The integrating factor $IF = e^{\int P(x) dx} = e^{-\int \frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x}$.
The solution is $t \cdot IF = \int Q(x) \cdot IF dx + C$.
$t \cdot \frac{1}{x} = \int (-\frac{1}{x^2}) \cdot \frac{1}{x} dx + C = -\int x^{-3} dx + C = -(\frac{x^{-2}}{-2}) + C = \frac{1}{2x^2} + C$.
Multiplying by $x$,we get $t = \frac{1}{2x} + Cx$.
Substituting $t = \frac{1}{\ln y}$,we get $\frac{1}{\ln y} = \frac{1}{2x} + Cx$.

(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 2$ and $Q = f(x)$.
Case $1$: For $x \in [0, 1]$,$f(x) = 1$.
$\frac{dy}{dx} + 2y = 1$. The integrating factor is $IF = e^{\int 2 dx} = e^{2x}$.
The solution is $y \cdot e^{2x} = \int 1 \cdot e^{2x} dx + C_1 = \frac{1}{2}e^{2x} + C_1$.
So,$y(x) = \frac{1}{2} + C_1 e^{-2x}$.
Given $y(0) = 0$,we have $0 = \frac{1}{2} + C_1 \Rightarrow C_1 = -\frac{1}{2}$.
Thus,$y(x) = \frac{1}{2} - \frac{1}{2}e^{-2x}$ for $x \in [0, 1]$.
At $x = 1$,$y(1) = \frac{1}{2} - \frac{1}{2}e^{-2} = \frac{e^2 - 1}{2e^2}$.
Case $2$: For $x > 1$,$f(x) = 0$.
$\frac{dy}{dx} + 2y = 0 \Rightarrow \frac{dy}{y} = -2 dx$.
Integrating both sides,$\ln|y| = -2x + C_2 \Rightarrow y = C_3 e^{-2x}$.
Using the continuity of $y(x)$ at $x = 1$,$y(1) = C_3 e^{-2} = \frac{e^2 - 1}{2e^2}$.
$C_3 = \frac{e^2 - 1}{2e^2} \cdot e^2 = \frac{e^2 - 1}{2}$.
So,$y(x) = \left(\frac{e^2 - 1}{2}\right) e^{-2x}$ for $x > 1$.
For $x = \frac{3}{2}$,$y\left(\frac{3}{2}\right) = \left(\frac{e^2 - 1}{2}\right) e^{-2(\frac{3}{2})} = \left(\frac{e^2 - 1}{2}\right) e^{-3} = \frac{e^2 - 1}{2e^3}$.

(D) Given the limit: $\mathop {\lim }\limits_{t \to x} \frac{{{t^2}f(x) - {x^2}f(t)}}{{t - x}} = 1$.
Applying $L$'Hopital's rule with respect to $t$:
$\mathop {\lim }\limits_{t \to x} \frac{2t f(x) - x^2 f'(t)}{1} = 1$.
Substituting $t = x$,we get the differential equation: $2x f(x) - x^2 f'(x) = 1$.
Rearranging: $f'(x) - \frac{2}{x} f(x) = -\frac{1}{x^2}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{2}{x}$ and $Q(x) = -\frac{1}{x^2}$.
The integrating factor $I.F. = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = \frac{1}{x^2}$.
Multiplying by $I.F.$: $\frac{d}{dx} [f(x) \cdot \frac{1}{x^2}] = -\frac{1}{x^4}$.
Integrating both sides: $\frac{f(x)}{x^2} = \int -x^{-4} dx = \frac{1}{3x^3} + C$.
So,$f(x) = \frac{1}{3x} + Cx^2$.
Using $f(1) = 1$: $1 = \frac{1}{3} + C \implies C = \frac{2}{3}$.
Thus,$f(x) = \frac{1}{3x} + \frac{2x^2}{3}$.
For $x = \frac{3}{2}$: $f(\frac{3}{2}) = \frac{1}{3(3/2)} + \frac{2(3/2)^2}{3} = \frac{2}{9} + \frac{2}{3} \cdot \frac{9}{4} = \frac{2}{9} + \frac{3}{2} = \frac{4 + 27}{18} = \frac{31}{18}$.

(D) Given the equation: $x \int_{1}^{x} y(t) dt = (x + 1) \int_{1}^{x} t y(t) dt$.
Differentiating both sides with respect to $x$ using the Leibniz rule:
$\int_{1}^{x} y(t) dt + x y(x) = \int_{1}^{x} t y(t) dt + (x + 1) x y(x)$.
Rearranging the terms:
$\int_{1}^{x} y(t) dt - \int_{1}^{x} t y(t) dt = (x^2 + x - x) y(x) = x^2 y(x)$.
Differentiating again with respect to $x$:
$y(x) - x y(x) = 2x y(x) + x^2 y'(x)$.
Simplifying:
$y(x) (1 - x - 2x) = x^2 y'(x) \implies y(x) (1 - 3x) = x^2 y'(x)$.
Separating variables:
$\frac{y'(x)}{y(x)} = \frac{1 - 3x}{x^2} = \frac{1}{x^2} - \frac{3}{x}$.
Integrating both sides:
$\ln|y(x)| = -\frac{1}{x} - 3 \ln|x| + K$.
$\ln|y(x)| + \ln|x^3| = -\frac{1}{x} + K$.
$\ln|y(x) x^3| = -\frac{1}{x} + K$.
$y(x) x^3 = C e^{-\frac{1}{x}}$.
Thus,$y(x) = \frac{C e^{-\frac{1}{x}}}{x^3}$.

(D) Given the differential equation: $\frac{dy}{dx} + \frac{y}{2} \sec x = \frac{\tan x}{2y}$.
Multiply by $2y$: $2y \frac{dy}{dx} + y^2 \sec x = \tan x$.
Let $y^2 = t$,then $2y \frac{dy}{dx} = \frac{dt}{dx}$.
The equation becomes: $\frac{dt}{dx} + t \sec x = \tan x$.
This is a linear differential equation of the form $\frac{dt}{dx} + Pt = Q$,where $P = \sec x$ and $Q = \tan x$.
Integrating factor $IF = e^{\int \sec x dx} = e^{\ln(\sec x + \tan x)} = \sec x + \tan x$.
The solution is $t(IF) = \int Q(IF) dx + C$.
$t(\sec x + \tan x) = \int \tan x(\sec x + \tan x) dx + C$.
$t(\sec x + \tan x) = \int (\sec x \tan x + \tan^2 x) dx + C$.
$t(\sec x + \tan x) = \int (\sec x \tan x + \sec^2 x - 1) dx + C$.
$t(\sec x + \tan x) = \sec x + \tan x - x + C$.
Given $y(0) = 1$,then $t(0) = 1^2 = 1$. At $x=0$,$\sec 0 = 1, \tan 0 = 0$.
$1(1 + 0) = 1 + 0 - 0 + C \Rightarrow C = 0$.
Thus,$t(\sec x + \tan x) = \sec x + \tan x - x$.
$t = 1 - \frac{x}{\sec x + \tan x}$.
Since $t = y^2$,we have $y^2 = 1 - \frac{x}{\sec x + \tan x}$.

(B) Given differential equation is $ydx - (x + 2y^2)dy = 0$.
Rearranging the terms,we get $ydx - xdy = 2y^2 dy$.
Dividing both sides by $y^2$ (for $y \neq 0$),we get $\frac{ydx - xdy}{y^2} = 2dy$.
This is equivalent to $d(\frac{x}{y}) = 2dy$.
Integrating both sides,we get $\frac{x}{y} = 2y + c$.
Given $f(-1) = 1$,which means $x = 1$ when $y = -1$. Substituting these values: $\frac{1}{-1} = 2(-1) + c \Rightarrow -1 = -2 + c \Rightarrow c = 1$.
Thus,the solution is $\frac{x}{y} = 2y + 1$,or $x = 2y^2 + y$.
To find $f(1)$,we substitute $y = 1$ into the equation: $x = 2(1)^2 + 1 = 2 + 1 = 3$.
Therefore,$f(1) = 3$.

(D) Given,$\sin 2x \left( \frac{dy}{dx} - \sqrt{\tan x} \right) - y = 0$
Rearranging the terms: $\frac{dy}{dx} - \frac{y}{\sin 2x} = \sqrt{\tan x}$
Since $\frac{1}{\sin 2x} = \csc 2x$,we have: $\frac{dy}{dx} - y \csc 2x = \sqrt{\tan x}$ ....$(1)$
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\csc 2x$ and $Q = \sqrt{\tan x}$.
Integrating Factor ($I$.$F$.) $= e^{\int P dx} = e^{\int -\csc 2x dx} = e^{-\frac{1}{2} \ln|\tan x|} = e^{\ln(\tan x)^{-1/2}} = \frac{1}{\sqrt{\tan x}} = \sqrt{\cot x}$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
Substituting the values: $y \sqrt{\cot x} = \int \sqrt{\tan x} \cdot \sqrt{\cot x} dx + c$.
Since $\sqrt{\tan x} \cdot \sqrt{\cot x} = 1$,we get: $y \sqrt{\cot x} = \int 1 dx + c$.
Therefore,$y \sqrt{\cot x} = x + c$.

(C) The given differential equation is $\frac{dy}{dx} + y \tan x = \sin 2x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \tan x$ and $Q = \sin 2x$.
The integrating factor $(IF)$ is $e^{\int P dx} = e^{\int \tan x dx} = e^{\ln(\sec x)} = \sec x$.
The general solution is $y(IF) = \int Q(IF) dx + c$.
$y \sec x = \int \sin 2x \sec x dx + c$.
$y \sec x = \int (2 \sin x \cos x) \sec x dx + c$.
$y \sec x = 2 \int \sin x dx + c$.
$y \sec x = -2 \cos x + c$ .....$(1)$.
Given $y(0) = 1$,substitute $x = 0$ and $y = 1$ into $(1)$:
$1 \cdot \sec(0) = -2 \cos(0) + c \Rightarrow 1(1) = -2(1) + c \Rightarrow c = 3$.
Substituting $c = 3$ into $(1)$,we get $y \sec x = -2 \cos x + 3$.
To find $y(\pi)$,substitute $x = \pi$:
$y \sec(\pi) = -2 \cos(\pi) + 3$.
$y(-1) = -2(-1) + 3$.
$-y = 2 + 3 = 5$.
$y = -5$.

(D) Given differential equation is $(1 + x^2) \frac{dy}{dx} + 2xy = 4x^2$.
Dividing by $(1 + x^2)$,we get $\frac{dy}{dx} + \frac{2x}{1 + x^2}y = \frac{4x^2}{1 + x^2}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2x}{1 + x^2}$ and $Q = \frac{4x^2}{1 + x^2}$.
Integrating factor $(I.F.) = e^{\int P dx} = e^{\int \frac{2x}{1 + x^2} dx} = e^{\ln(1 + x^2)} = 1 + x^2$.
The general solution is $y \times (I.F.) = \int Q \times (I.F.) dx + C$.
$y(1 + x^2) = \int \frac{4x^2}{1 + x^2} \times (1 + x^2) dx + C$.
$y(1 + x^2) = \int 4x^2 dx + C$.
$y(1 + x^2) = \frac{4x^3}{3} + C$.
Since the curve passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ to find $C$.
$0(1 + 0) = \frac{4(0)^3}{3} + C \implies C = 0$.
Thus,the equation of the curve is $y(1 + x^2) = \frac{4x^3}{3}$,which simplifies to $3(1 + x^2)y = 4x^3$.

(D) The given differential equation is $\frac{dy}{dx} + \frac{2}{x}y = x^2$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2}{x}$ and $Q = x^2$.
First,we calculate the integrating factor ($I$.$F$.):
$I.F. = e^{\int P dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln|x|} = e^{\ln(x^2)} = x^2$.
The general solution is given by $y \cdot (I.F.) = \int (Q \cdot I.F.) dx + c$.
Substituting the values,we get $y \cdot x^2 = \int (x^2 \cdot x^2) dx + c$.
$y \cdot x^2 = \int x^4 dx + c$.
$y \cdot x^2 = \frac{x^5}{5} + c$.
Dividing both sides by $x^2$,we get $y = \frac{x^3}{5} + cx^{-2}$.

(B) The given differential equation is $(x^2 - 1)\frac{dy}{dx} + 2xy = x$.
Dividing both sides by $(x^2 - 1)$,we get $\frac{dy}{dx} + \frac{2x}{x^2 - 1}y = \frac{x}{x^2 - 1}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2x}{x^2 - 1}$ and $Q = \frac{x}{x^2 - 1}$.
The integrating factor $(IF)$ is given by $e^{\int P dx} = e^{\int \frac{2x}{x^2 - 1} dx}$.
Let $t = x^2 - 1$,then $dt = 2x dx$.
Thus,$IF = e^{\int \frac{dt}{t}} = e^{\ln|t|} = t = x^2 - 1$.
Therefore,the integrating factor is $x^2 - 1$.

(A) The given differential equation is $\frac{dy}{dx} + (3 \sec^2 x) y = \sec^2 x$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x) y = Q(x)$,where $P(x) = 3 \sec^2 x$ and $Q(x) = \sec^2 x$.
The integrating factor $(IF)$ is $e^{\int P(x) dx} = e^{\int 3 \sec^2 x dx} = e^{3 \tan x}$.
The general solution is $y \cdot IF = \int Q(x) \cdot IF dx + C$.
$y \cdot e^{3 \tan x} = \int \sec^2 x \cdot e^{3 \tan x} dx + C$.
Let $u = 3 \tan x$,then $du = 3 \sec^2 x dx$,so $\sec^2 x dx = \frac{du}{3}$.
$y \cdot e^{3 \tan x} = \int e^u \cdot \frac{du}{3} + C = \frac{1}{3} e^{3 \tan x} + C$.
Dividing by $e^{3 \tan x}$,we get $y = \frac{1}{3} + C e^{-3 \tan x}$.
Given $y\left( \frac{\pi}{4} \right) = \frac{4}{3}$,we have $\frac{4}{3} = \frac{1}{3} + C e^{-3 \tan(\pi/4)} = \frac{1}{3} + C e^{-3}$.
$1 = C e^{-3} \Rightarrow C = e^3$.
Thus,$y(x) = \frac{1}{3} + e^3 \cdot e^{-3 \tan x} = \frac{1}{3} + e^{3 - 3 \tan x}$.
For $x = -\frac{\pi}{4}$,$y\left( -\frac{\pi}{4} \right) = \frac{1}{3} + e^{3 - 3 \tan(-\pi/4)} = \frac{1}{3} + e^{3 - 3(-1)} = \frac{1}{3} + e^6$.

(B) The given differential equation is $f'(x) + \frac{3}{4x} f(x) = 7$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{3}{4x}$ and $Q(x) = 7$.
The integrating factor $IF = e^{\int P(x) dx} = e^{\int \frac{3}{4x} dx} = e^{\frac{3}{4} \ln x} = x^{3/4}$.
The general solution is $f(x) \cdot IF = \int Q(x) \cdot IF dx + c$.
$f(x) \cdot x^{3/4} = \int 7 x^{3/4} dx + c = 7 \cdot \frac{x^{7/4}}{7/4} + c = 4x^{7/4} + c$.
Thus,$f(x) = 4x + c x^{-3/4}$.
Now,we evaluate $\lim_{x \to 0^+} x f\left(\frac{1}{x}\right)$.
$f\left(\frac{1}{x}\right) = 4\left(\frac{1}{x}\right) + c\left(\frac{1}{x}\right)^{-3/4} = \frac{4}{x} + c x^{3/4}$.
Therefore,$\lim_{x \to 0^+} x f\left(\frac{1}{x}\right) = \lim_{x \to 0^+} x \left(\frac{4}{x} + c x^{3/4}\right) = \lim_{x \to 0^+} (4 + c x^{7/4}) = 4$.

(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = 2 + \frac{1}{x}$ and $Q(x) = e^{-2x}$.
The Integrating Factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int (2 + \frac{1}{x}) dx} = e^{2x + \log_e x} = x e^{2x}$.
The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.
$y(x e^{2x}) = \int e^{-2x} \cdot (x e^{2x}) dx + C = \int x dx + C = \frac{x^2}{2} + C$.
Given $y(1) = \frac{1}{2}e^{-2}$,we substitute $x=1$ and $y=\frac{1}{2}e^{-2}$:
$\frac{1}{2}e^{-2} \cdot (1 \cdot e^2) = \frac{1^2}{2} + C \Rightarrow \frac{1}{2} = \frac{1}{2} + C \Rightarrow C = 0$.
Thus,$y = \frac{x}{2}e^{-2x}$.
To check the monotonicity,we find $\frac{dy}{dx} = \frac{1}{2} e^{-2x} + \frac{x}{2} (-2 e^{-2x}) = \frac{e^{-2x}}{2} (1 - 2x)$.
For $x \in \left( \frac{1}{2}, 1 \right)$,$1 - 2x < 0$,so $\frac{dy}{dx} < 0$,which means $y(x)$ is decreasing in $\left( \frac{1}{2}, 1 \right)$.

(C) The given differential equation is $x\frac{dy}{dx} + y = x \ln x$.
Dividing by $x$,we get $\frac{dy}{dx} + \frac{1}{x}y = \ln x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = \ln x$.
The integrating factor $IF = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.
The general solution is $y \cdot IF = \int Q \cdot IF dx + C$.
$y \cdot x = \int (\ln x) \cdot x dx + C$.
Using integration by parts,$\int x \ln x dx = \ln x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$.
So,$xy = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$.
Given $2y(2) = \ln 4 - 1$,we have $y(2) = \frac{\ln 4 - 1}{2}$.
Substituting $x=2$ in the general solution: $2y(2) = \frac{2^2}{2} \ln 2 - \frac{2^2}{4} + C \implies \ln 4 - 1 = 2 \ln 2 - 1 + C$.
Since $\ln 4 = 2 \ln 2$,we get $2 \ln 2 - 1 = 2 \ln 2 - 1 + C$,which implies $C = 0$.
Thus,$y = \frac{x}{2} \ln x - \frac{x}{4}$.
For $x = e$,$y(e) = \frac{e}{2} \ln e - \frac{e}{4} = \frac{e}{2} - \frac{e}{4} = \frac{e}{4}$.

(B) The slope of the tangent is given by $\frac{dy}{dx} = \frac{x^2 - 2y}{x}$.
This can be rewritten as a linear differential equation: $\frac{dy}{dx} + \frac{2}{x}y = x$.
This is of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2}{x}$ and $Q = x$.
The integrating factor $IF = e^{\int P dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln|x|} = x^2$.
The general solution is $y \cdot IF = \int Q \cdot IF dx + C$.
$y \cdot x^2 = \int x \cdot x^2 dx + C = \int x^3 dx + C = \frac{x^4}{4} + C$.
Since the curve passes through $(1, -2)$,we substitute $x = 1$ and $y = -2$:
$-2(1)^2 = \frac{1^4}{4} + C \Rightarrow -2 = \frac{1}{4} + C \Rightarrow C = -2 - \frac{1}{4} = -\frac{9}{4}$.
Thus,the equation of the curve is $y x^2 = \frac{x^4}{4} - \frac{9}{4}$,or $4yx^2 = x^4 - 9$.
Checking option $(B)$: For $x = \sqrt{3}$,$4y(3) = (\sqrt{3})^4 - 9 = 9 - 9 = 0$,so $y = 0$. Thus,the curve passes through $(\sqrt{3}, 0)$.

(C) The given differential equation is $(x^2 + 1)^2 \frac{dy}{dx} + 2x(x^2 + 1)y = 1$.
Dividing by $(x^2 + 1)^2$,we get $\frac{dy}{dx} + \frac{2x}{x^2 + 1} y = \frac{1}{(x^2 + 1)^2}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{2x}{x^2 + 1}$ and $Q(x) = \frac{1}{(x^2 + 1)^2}$.
The integrating factor $I.F. = e^{\int P(x) dx} = e^{\int \frac{2x}{x^2 + 1} dx} = e^{\ln(x^2 + 1)} = x^2 + 1$.
The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.
$y(x^2 + 1) = \int \frac{1}{(x^2 + 1)^2} \cdot (x^2 + 1) dx + C = \int \frac{1}{x^2 + 1} dx + C$.
$y(x^2 + 1) = \tan^{-1}(x) + C$.
Given $y(0) = 0$,we have $0(0^2 + 1) = \tan^{-1}(0) + C$,which implies $C = 0$.
Thus,$y(x) = \frac{\tan^{-1}(x)}{x^2 + 1}$.
We are given $\sqrt{a} y(1) = \frac{\pi}{32}$.
$y(1) = \frac{\tan^{-1}(1)}{1^2 + 1} = \frac{\pi/4}{2} = \frac{\pi}{8}$.
Substituting this into the equation: $\sqrt{a} \cdot \frac{\pi}{8} = \frac{\pi}{32}$.
$\sqrt{a} = \frac{8}{32} = \frac{1}{4}$.
Therefore,$a = (1/4)^2 = 1/16$.

(B) Given the differential equation: $x \frac{dy}{dx} + 2y = x^2$.
Divide by $x$ to get the standard form of a linear differential equation: $\frac{dy}{dx} + \frac{2}{x}y = x$.
Here,$P(x) = \frac{2}{x}$ and $Q(x) = x$.
The Integrating Factor $(IF)$ is given by: $IF = e^{\int P(x) dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln|x|} = x^2$.
The general solution is: $y \cdot IF = \int Q(x) \cdot IF dx + C$.
$y \cdot x^2 = \int x \cdot x^2 dx = \int x^3 dx = \frac{x^4}{4} + C$.
Using the condition $y(1) = 1$:
$1 \cdot (1)^2 = \frac{1^4}{4} + C \Rightarrow 1 = \frac{1}{4} + C \Rightarrow C = \frac{3}{4}$.
Substituting $C$ back into the general solution:
$y x^2 = \frac{x^4}{4} + \frac{3}{4}$.
Dividing by $x^2$,we get: $y = \frac{x^2}{4} + \frac{3}{4x^2}$.

(D) The given differential equation is $\cos x \frac{dy}{dx} - y \sin x = 6x$.
Dividing by $\cos x$,we get $\frac{dy}{dx} - y \tan x = 6x \sec x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\tan x$ and $Q = 6x \sec x$.
The integrating factor is $IF = e^{\int P dx} = e^{-\int \tan x dx} = e^{-\ln(\sec x)} = \frac{1}{\sec x} = \cos x$.
The general solution is $y \cdot IF = \int Q \cdot IF dx + C$.
$y \cos x = \int (6x \sec x) \cdot \cos x dx = \int 6x dx = 3x^2 + C$.
Given $y(\frac{\pi}{3}) = 0$,we have $0 \cdot \cos(\frac{\pi}{3}) = 3(\frac{\pi}{3})^2 + C$,which gives $0 = \frac{\pi^2}{3} + C$,so $C = -\frac{\pi^2}{3}$.
Thus,$y \cos x = 3x^2 - \frac{\pi^2}{3}$.
At $x = \frac{\pi}{6}$,$\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$.
$y \cdot \frac{\sqrt{3}}{2} = 3(\frac{\pi}{6})^2 - \frac{\pi^2}{3} = 3(\frac{\pi^2}{36}) - \frac{\pi^2}{3} = \frac{\pi^2}{12} - \frac{4\pi^2}{12} = -\frac{3\pi^2}{12} = -\frac{\pi^2}{4}$.
$y = -\frac{\pi^2}{4} \cdot \frac{2}{\sqrt{3}} = -\frac{\pi^2}{2\sqrt{3}}$.

(C) The given differential equation is $\frac{dy}{dx} + y \sec^2 x = \tan x \sec^2 x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \sec^2 x$ and $Q = \tan x \sec^2 x$.
The integrating factor is $IF = e^{\int \sec^2 x dx} = e^{\tan x}$.
The solution is $y \cdot e^{\tan x} = \int (\tan x \sec^2 x) e^{\tan x} dx + C$.
Let $t = \tan x$,then $dt = \sec^2 x dx$.
The integral becomes $\int t e^t dt = t e^t - e^t + C$.
So,$y e^{\tan x} = e^{\tan x} (\tan x - 1) + C$.
Given $y(0) = 0$,we have $0 = e^0 (0 - 1) + C$,which implies $C = 1$.
Thus,$y = \tan x - 1 + e^{-\tan x}$.
For $x = -\frac{\pi}{4}$,$\tan x = -1$.
$y\left( -\frac{\pi}{4} \right) = -1 - 1 + e^{-(-1)} = e - 2$.

(B) The given differential equation 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 ($I$.$F$.) is $e^{\int P dx} = e^{\int \tan x dx} = e^{\ln |\sec x|} = \sec x$.
Multiplying both sides by the $I$.$F$.,we get $y \sec x = \int (2x + x^2 \tan x) \sec x dx$.
$y \sec x = \int 2x \sec x dx + \int x^2 \sec x \tan x dx$.
Using integration by parts for the first integral: $\int 2x \sec x dx$ is complex,but notice that $\frac{d}{dx}(x^2 \sec x) = 2x \sec x + x^2 \sec x \tan x$.
Thus,$y \sec x = x^2 \sec x + C$.
Dividing by $\sec x$,we get $y = x^2 + C \cos x$.
Given $y(0) = 1$,we have $1 = 0^2 + C \cos(0) \Rightarrow C = 1$.
So,$y = x^2 + \cos x$.
Now,$y' = 2x - \sin x$.
$y'\left( \frac{\pi}{4} \right) = 2\left( \frac{\pi}{4} \right) - \sin\left( \frac{\pi}{4} \right) = \frac{\pi}{2} - \frac{1}{\sqrt{2}}$.
$y'\left( -\frac{\pi}{4} \right) = 2\left( -\frac{\pi}{4} \right) - \sin\left( -\frac{\pi}{4} \right) = -\frac{\pi}{2} + \frac{1}{\sqrt{2}}$.
$y'\left( \frac{\pi}{4} \right) - y'\left( -\frac{\pi}{4} \right) = \left( \frac{\pi}{2} - \frac{1}{\sqrt{2}} \right) - \left( -\frac{\pi}{2} + \frac{1}{\sqrt{2}} \right) = \pi - \frac{2}{\sqrt{2}} = \pi - \sqrt{2}$.
Thus,option $B$ is correct.

(C) The given differential equation is $y^2 dx + (x - \frac{1}{y}) dy = 0$.
Rearranging the terms,we get $y^2 dx + x dy = \frac{1}{y} dy$.
Dividing by $y^2 dy$,we obtain $\frac{dx}{dy} + \frac{x}{y^2} = \frac{1}{y^3}$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{y^2}$ and $Q(y) = \frac{1}{y^3}$.
The integrating factor $(IF)$ is $e^{\int P(y) dy} = e^{\int \frac{1}{y^2} dy} = e^{-\frac{1}{y}}$.
The general solution is $x \cdot e^{-\frac{1}{y}} = \int \frac{1}{y^3} e^{-\frac{1}{y}} dy + C$.
Let $t = -\frac{1}{y}$,then $dt = \frac{1}{y^2} dy$. Also,$\frac{1}{y} = -t$.
Substituting these,$\int (-t) e^t dt = - (t e^t - e^t) = e^t(1 - t) = e^{-\frac{1}{y}}(1 + \frac{1}{y})$.
Thus,$x e^{-\frac{1}{y}} = e^{-\frac{1}{y}}(1 + \frac{1}{y}) + C$.
Given $y=1$ when $x=1$,we have $1 \cdot e^{-1} = e^{-1}(1 + 1) + C \implies e^{-1} = 2e^{-1} + C \implies C = -e^{-1}$.
So,$x e^{-\frac{1}{y}} = e^{-\frac{1}{y}}(1 + \frac{1}{y}) - e^{-1}$.
For $y=2$,$x e^{-1/2} = e^{-1/2}(1 + 1/2) - e^{-1} = \frac{3}{2} e^{-1/2} - e^{-1}$.
Dividing by $e^{-1/2}$,we get $x = \frac{3}{2} - e^{-1/2} = \frac{3}{2} - \frac{1}{\sqrt{e}}$.

(B) Given differential equation: $(y^2 - x^3) dx - xy dy = 0$
Divide by $dx$ (assuming $dx \neq 0$):
$y^2 - x^3 - xy \frac{dy}{dx} = 0$
$xy \frac{dy}{dx} - y^2 = -x^3$
Divide by $x$:
$y \frac{dy}{dx} - \frac{1}{x} y^2 = -x^2$ ......$(i)$
Let $y^2 = v$,then $2y \frac{dy}{dx} = \frac{dv}{dx}$,so $y \frac{dy}{dx} = \frac{1}{2} \frac{dv}{dx}$.
Substituting into $(i)$:
$\frac{1}{2} \frac{dv}{dx} - \frac{1}{x} v = -x^2$
$\frac{dv}{dx} - \frac{2}{x} v = -2x^2$ ......$(ii)$
This is a linear differential equation of the form $\frac{dv}{dx} + P(x)v = Q(x)$,where $P(x) = -\frac{2}{x}$ and $Q(x) = -2x^2$.
Integrating factor $I.F. = e^{\int P(x) dx} = e^{\int -\frac{2}{x} dx} = e^{-2 \ln |x|} = x^{-2} = \frac{1}{x^2}$.
The solution is $v \cdot I.F. = \int Q(x) \cdot I.F. dx + c$.
$v \cdot \frac{1}{x^2} = \int (-2x^2) \cdot \frac{1}{x^2} dx + c$
$\frac{v}{x^2} = \int -2 dx + c$
$\frac{v}{x^2} = -2x + c$
Since $v = y^2$,we have $\frac{y^2}{x^2} = -2x + c$.
$y^2 = -2x^3 + cx^2$,which implies $y^2 - 2x^3 + cx^2 = 0$.
Thus,the correct option is $B$.