Variable separable type differential equations Questions in English

(C) Given the differential equation $\frac{dy}{dx} = e^{-2y}$.
Separating the variables,we get $e^{2y} dy = dx$.
Integrating both sides,we have $\int e^{2y} dy = \int dx$,which gives $\frac{e^{2y}}{2} = x + C$.
Using the initial condition $y = 0$ when $x = 5$,we substitute these values into the equation: $\frac{e^{2(0)}}{2} = 5 + C$.
$\frac{1}{2} = 5 + C$,which implies $C = \frac{1}{2} - 5 = -\frac{9}{2}$.
So,the equation becomes $\frac{e^{2y}}{2} = x - \frac{9}{2}$.
Now,to find $x$ when $y = 3$,we substitute $y = 3$ into the equation: $\frac{e^{2(3)}}{2} = x - \frac{9}{2}$.
$\frac{e^6}{2} = x - \frac{9}{2}$,which gives $x = \frac{e^6}{2} + \frac{9}{2} = \frac{e^6 + 9}{2}$.

(A) Given the differential equation: $dy - \sin x \sin y \, dx = 0$.
Rearranging the terms to separate the variables: $dy = \sin x \sin y \, dx$.
Dividing both sides by $\sin y$: $\frac{dy}{\sin y} = \sin x \, dx$.
Integrating both sides: $\int \csc y \, dy = \int \sin x \, dx$.
Using the standard integral $\int \csc y \, dy = \ln |\tan \frac{y}{2}|$: $\ln |\tan \frac{y}{2}| = -\cos x + C$.
Taking the exponential of both sides: $\tan \frac{y}{2} = e^{-\cos x + C} = e^{-\cos x} \cdot e^C$.
Let $e^C = c_1$,then $\tan \frac{y}{2} = c_1 e^{-\cos x}$.
Multiplying by $e^{\cos x}$: $e^{\cos x} \tan \frac{y}{2} = c_1$.

(C) Given the differential equation: ${e^y}\frac{{dy}}{{dx}} + ({e^y} + 1)\cot x = 0$
Rearranging the terms to separate the variables:
${e^y}\frac{{dy}}{{dx}} = -({e^y} + 1)\cot x$
$\frac{{{e^y}}}{{{e^y} + 1}}dy = -\cot x dx$
Integrating both sides:
$\int \frac{{{e^y}}}{{{e^y} + 1}}dy = -\int \cot x dx$
Let $u = {e^y} + 1$,then $du = {e^y} dy$. The integral becomes:
$\int \frac{1}{u} du = -\int \cot x dx$
$\ln|{e^y} + 1| = -\ln|\sin x| + C$
$\ln|{e^y} + 1| + \ln|\sin x| = C$
Using the property $\ln A + \ln B = \ln(AB)$:
$\ln|({e^y} + 1)\sin x| = C$
Taking the exponential of both sides,we get:
$({e^y} + 1)\sin x = K$ (where $K = e^C$ is a constant).

(A) Given the differential equation: $\frac{dy}{dx} = \sin x + 2x$.
Integrating both sides with respect to $x$:
$\int dy = \int (\sin x + 2x) dx$.
Using the standard integrals $\int \sin x dx = -\cos x$ and $\int x^n dx = \frac{x^{n+1}}{n+1}$:
$y = -\cos x + 2 \cdot \frac{x^2}{2} + c$.
Simplifying the expression:
$y = x^2 - \cos x + c$.

(A) Given the differential equation: $\frac{dy}{dx} = 2xy$.
Separate the variables: $\frac{dy}{y} = 2x \, dx$.
Integrate both sides: $\int \frac{dy}{y} = \int 2x \, dx$.
This gives: $\ln|y| = x^2 + C_1$.
Exponentiate both sides: $|y| = e^{x^2 + C_1} = e^{C_1} e^{x^2}$.
Let $C = \pm e^{C_1}$,then the general solution is $y = Ce^{x^2}$.

(C) Given the differential equation: $ydx - xdy = x^2 ydx$.
Rearranging the terms,we get: $-xdy = x^2 ydx - ydx$.
$-xdy = y(x^2 - 1)dx$.
Separating the variables,we have: $\frac{dy}{y} = \frac{x^2 - 1}{-x} dx = (\frac{1}{x} - x) dx$.
Integrating both sides: $\int \frac{dy}{y} = \int (\frac{1}{x} - x) dx$.
$\ln|y| = \ln|x| - \frac{x^2}{2} + C$.
Multiplying by $2$: $2 \ln|y| = 2 \ln|x| - x^2 + 2C$.
$\ln(y^2) = \ln(x^2) - x^2 + K$ (where $K = 2C$).
$\ln(y^2) - \ln(x^2) = -x^2 + K$.
$\ln(\frac{y^2}{x^2}) = -x^2 + K$.
Taking the exponential of both sides: $\frac{y^2}{x^2} = e^{-x^2 + K} = e^K e^{-x^2}$.
Let $e^K = \frac{1}{c}$,then $y^2 = \frac{x^2}{c} e^{-x^2}$,which implies $c y^2 = x^2 e^{-x^2}$ or $y^2 e^{x^2} = c x^2$ (adjusting the constant $c$ appropriately).
Thus,the correct option is $(c)$.

(D) To solve the differential equation $\frac{dy}{dx} = 4x + y + 1$,we observe that the right-hand side is a linear function of $x$ and $y$.
Let $v = 4x + y + 1$.
Differentiating both sides with respect to $x$,we get $\frac{dv}{dx} = 4 + \frac{dy}{dx}$.
Thus,$\frac{dy}{dx} = \frac{dv}{dx} - 4$.
Substituting these into the original equation,we get $\frac{dv}{dx} - 4 = v$,which is a variable separable differential equation.
Therefore,the suitable substitution is $y + 4x + 1 = v$.

(C) Given equation is $(x + y - 1)dx + (2x + 2y - 3)dy = 0$,which can be written as $\frac{dy}{dx} = - \frac{x + y - 1}{2(x + y) - 3}$.
Let $x + y = t$. Then $1 + \frac{dy}{dx} = \frac{dt}{dx}$,so $\frac{dy}{dx} = \frac{dt}{dx} - 1$.
Substituting these into the equation: $\frac{dt}{dx} - 1 = - \frac{t - 1}{2t - 3}$.
$\frac{dt}{dx} = 1 - \frac{t - 1}{2t - 3} = \frac{2t - 3 - t + 1}{2t - 3} = \frac{t - 2}{2t - 3}$.
Separating the variables: $\frac{2t - 3}{t - 2} dt = dx$.
$\frac{2(t - 2) + 1}{t - 2} dt = dx \implies (2 + \frac{1}{t - 2}) dt = dx$.
Integrating both sides: $\int (2 + \frac{1}{t - 2}) dt = \int dx$.
$2t + \log |t - 2| = x + c$.
Substituting $t = x + y$: $2(x + y) + \log |x + y - 2| = x + c$.
$2x + 2y + \log |x + y - 2| = x + c \implies x + 2y + \log |x + y - 2| = c$.

(B) Given the differential equation $\sin \left( \frac{dy}{dx} \right) = a$.
Taking the inverse sine on both sides,we get $\frac{dy}{dx} = \sin^{-1}(a)$.
This is a variable separable differential equation.
Integrating both sides with respect to $x$,we have $\int dy = \int \sin^{-1}(a) \, dx$.
Since $\sin^{-1}(a)$ is a constant,we get $y = x \sin^{-1}(a) + c$.
Using the initial condition $y(0) = 1$,we substitute $x = 0$ and $y = 1$:
$1 = 0 \cdot \sin^{-1}(a) + c \implies c = 1$.
Thus,the equation becomes $y = x \sin^{-1}(a) + 1$.
Rearranging the terms,$y - 1 = x \sin^{-1}(a)$,which implies $\frac{y - 1}{x} = \sin^{-1}(a)$.
Taking the sine of both sides,we obtain $\sin \left( \frac{y - 1}{x} \right) = a$.

(A) Given the differential equation: $\cos (x + y) \, dy = dx$ ... $(i)$
Let $x + y = v$. Differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dv}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$.
Substituting these into $(i)$,we have $\cos v \left( \frac{dv}{dx} - 1 \right) = 1$.
This simplifies to $\cos v \frac{dv}{dx} = 1 + \cos v$,or $\frac{\cos v}{1 + \cos v} \, dv = dx$.
Using the identity $\cos v = 2 \cos^2(v/2) - 1$,we get $\frac{2 \cos^2(v/2) - 1}{2 \cos^2(v/2)} \, dv = dx$,which simplifies to $\left( 1 - \frac{1}{2} \sec^2(v/2) \right) \, dv = dx$.
Integrating both sides,we get $v - \tan(v/2) = x + c$.
Substituting $v = x + y$ back,we get $(x + y) - \tan \left( \frac{x + y}{2} \right) = x + c$,which simplifies to $y = \tan \left( \frac{x + y}{2} \right) + c$.

(A) Given differential equation is $\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1 - x^2}} = 0$.
Rearranging the terms,we get $\frac{dy}{dx} = -\frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}}$.
Separating the variables,we have $\frac{dy}{\sqrt{1 - y^2}} = -\frac{dx}{\sqrt{1 - x^2}}$.
Integrating both sides,$\int \frac{dy}{\sqrt{1 - y^2}} = -\int \frac{dx}{\sqrt{1 - x^2}}$.
This gives $\sin^{-1} y = -\sin^{-1} x + c$.
Rearranging the terms,we get $\sin^{-1} x + \sin^{-1} y = c$.

(C) Given the differential equation $\frac{dy}{dx} = 2^{y - x}$.
Using the property of exponents,we can write this as $\frac{dy}{dx} = \frac{2^y}{2^x}$.
Separating the variables,we get $\frac{dy}{2^y} = \frac{dx}{2^x}$,which is $2^{-y} dy = 2^{-x} dx$.
Integrating both sides,we have $\int 2^{-y} dy = \int 2^{-x} dx$.
Using the formula $\int a^u du = \frac{a^u}{\ln a}$,we get $\frac{2^{-y}}{-\ln 2} = \frac{2^{-x}}{-\ln 2} + C_1$.
Multiplying by $-\ln 2$,we get $2^{-y} = 2^{-x} - C_1 \ln 2$.
Rearranging the terms,$2^{-x} - 2^{-y} = C_1 \ln 2$.
Let $C = C_1 \ln 2$,then $\frac{1}{2^x} - \frac{1}{2^y} = c$.

(B) Given differential equation is $\frac{dy}{dx} + \sin \left( \frac{x + y}{2} \right) = \sin \left( \frac{x - y}{2} \right)$.
Rearranging the terms,we get $\frac{dy}{dx} = \sin \left( \frac{x - y}{2} \right) - \sin \left( \frac{x + y}{2} \right)$.
Using the trigonometric identity $\sin A - \sin B = 2 \cos \left( \frac{A + B}{2} \right) \sin \left( \frac{A - B}{2} \right)$,we have $\sin \left( \frac{x - y}{2} \right) - \sin \left( \frac{x + y}{2} \right) = 2 \cos \left( \frac{x}{2} \right) \sin \left( -\frac{y}{2} \right) = -2 \cos \left( \frac{x}{2} \right) \sin \left( \frac{y}{2} \right)$.
Thus,$\frac{dy}{dx} = -2 \cos \left( \frac{x}{2} \right) \sin \left( \frac{y}{2} \right)$.
Separating the variables,we get $\text{cosec} \left( \frac{y}{2} \right) dy = -2 \cos \left( \frac{x}{2} \right) dx$.
Integrating both sides,$\int \text{cosec} \left( \frac{y}{2} \right) dy = -2 \int \cos \left( \frac{x}{2} \right) dx + c$.
Using $\int \text{cosec} (ax) dx = \frac{1}{a} \log \tan \left( \frac{ax}{2} \right)$,we get $\frac{\log \tan (y/4)}{1/2} = -2 \frac{\sin (x/2)}{1/2} + c$.
This simplifies to $2 \log \tan (y/4) = -4 \sin (x/2) + c$,or $\log \tan (y/4) = c - 2 \sin (x/2)$.

(D) Given differential equation is $(x + y)^2 \frac{dy}{dx} = a^2$.
Let $x + y = v$.
Differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dv}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$.
Substituting these into the original equation:
$v^2 (\frac{dv}{dx} - 1) = a^2$
$\frac{dv}{dx} - 1 = \frac{a^2}{v^2}$
$\frac{dv}{dx} = \frac{a^2}{v^2} + 1 = \frac{a^2 + v^2}{v^2}$
Separating the variables:
$\frac{v^2}{a^2 + v^2} dv = dx$
Integrating both sides:
$\int \frac{v^2 + a^2 - a^2}{a^2 + v^2} dv = \int dx$
$\int (1 - \frac{a^2}{a^2 + v^2}) dv = x + c$
$v - a \tan^{-1}(\frac{v}{a}) = x + c$
Substituting $v = x + y$ back:
$(x + y) - a \tan^{-1}(\frac{x + y}{a}) = x + c$
$y - a \tan^{-1}(\frac{x + y}{a}) = c$
Thus,the correct option is $(d)$.

(A) Given the differential equation: $\frac{dx}{x} + \frac{dy}{y} = 0$.
Integrating both sides,we get: $\int \frac{dx}{x} + \int \frac{dy}{y} = \int 0$.
This results in: $\log |x| + \log |y| = C_1$.
Using the logarithmic property $\log a + \log b = \log(ab)$,we have: $\log |xy| = C_1$.
Taking the exponential of both sides,we get $|xy| = e^{C_1}$.
Letting $e^{C_1} = c$,we obtain the final solution: $xy = c$.

(B) Given the differential equation: $y - x\frac{dy}{dx} = a(y^2 + \frac{dy}{dx})$
Rearranging the terms to separate variables:
$y - ay^2 = a\frac{dy}{dx} + x\frac{dy}{dx}$
$y(1 - ay) = (a + x)\frac{dy}{dx}$
Separating the variables:
$\frac{dx}{a + x} = \frac{dy}{y(1 - ay)}$
Using partial fractions for the right side:
$\frac{1}{y(1 - ay)} = \frac{1}{y} + \frac{a}{1 - ay}$
Integrating both sides:
$\int \frac{dx}{a + x} = \int \left(\frac{1}{y} + \frac{a}{1 - ay}\right) dy$
$\ln|a + x| = \ln|y| - \ln|1 - ay| + \ln|c|$
$\ln|a + x| = \ln|\frac{cy}{1 - ay}|$
Taking the exponential of both sides:
$a + x = \frac{cy}{1 - ay}$
$(x + a)(1 - ay) = cy$

(B) Given the differential equation: $\log \left( \frac{dy}{dx} \right) = ax + by$
Taking the exponential of both sides,we get: $\frac{dy}{dx} = e^{ax + by}$
Using the property of exponents $e^{m+n} = e^m \cdot e^n$,we can write: $\frac{dy}{dx} = e^{ax} \cdot e^{by}$
Separating the variables $x$ and $y$: $e^{-by} \, dy = e^{ax} \, dx$
Integrating both sides: $\int e^{-by} \, dy = \int e^{ax} \, dx$
Performing the integration: $\frac{e^{-by}}{-b} = \frac{e^{ax}}{a} + c$
Thus,the solution is $\frac{e^{-by}}{-b} = \frac{e^{ax}}{a} + c$.

(C) Given the differential equation: $\frac{dy}{dx} = \left( \frac{y}{x} \right)^{1/3}$.
Separating the variables,we get: $\frac{dy}{y^{1/3}} = \frac{dx}{x^{1/3}}$.
Integrating both sides: $\int y^{-1/3} dy = \int x^{-1/3} dx$.
Using the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + c$,we get:
$\frac{y^{2/3}}{2/3} = \frac{x^{2/3}}{2/3} + c$.
Multiplying by $\frac{2}{3}$,we obtain: $y^{2/3} = x^{2/3} + c'$,where $c'$ is a constant.
Rearranging the terms,we get: $y^{2/3} - x^{2/3} = c$.

(C) Given the differential equation: $(2y - 1) \, dx = (2x + 3) \, dy$
Separating the variables,we get: $\frac{dx}{2x + 3} = \frac{dy}{2y - 1}$
Integrating both sides: $\int \frac{dx}{2x + 3} = \int \frac{dy}{2y - 1}$
$\frac{1}{2} \ln|2x + 3| = \frac{1}{2} \ln|2y - 1| + C'$
Multiply by $2$: $\ln|2x + 3| = \ln|2y - 1| + 2C'$
$\ln|2x + 3| - \ln|2y - 1| = \ln c$ (where $\ln c = 2C'$)
$\ln \left| \frac{2x + 3}{2y - 1} \right| = \ln c$
Therefore,$\frac{2x + 3}{2y - 1} = c$.

(B) Given the differential equation: $\cot y \, dx = x \, dy$
Separate the variables:
$\frac{dx}{x} = \frac{dy}{\cot y}$
Since $\frac{1}{\cot y} = \tan y$,we have:
$\frac{dx}{x} = \tan y \, dy$
Integrating both sides:
$\int \frac{dx}{x} = \int \tan y \, dy$
$\ln |x| = \ln |\sec y| + \ln |c|$
Using the property of logarithms $\ln a + \ln b = \ln(ab)$:
$\ln |x| = \ln |c \sec y|$
Taking the exponential of both sides:
$x = c \sec y$

(A) Given the differential equation: $\frac{dy}{dx} = \frac{x \log(x^2) + x}{\sin y + y \cos y}$.
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 $\int y \cos y dy$:
$\int \sin y dy + (y \sin y - \int \sin y dy) = y \sin y$.
For the right side,note that $\log(x^2) = 2 \log x$:
$\int (2x \log x + x) dx = 2 \int x \log x dx + \int x dx$.
Using integration by parts for $\int x \log x dx$:
$2 [\frac{x^2}{2} \log x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx] + \frac{x^2}{2} = 2 [\frac{x^2}{2} \log x - \frac{x^2}{4}] + \frac{x^2}{2} = x^2 \log x - \frac{x^2}{2} + \frac{x^2}{2} = x^2 \log x$.
Thus,the solution is $y \sin y = x^2 \log x + c$.

(C) The given differential equation is $\left( \frac{2 + \sin x}{1 + y} \right) \frac{dy}{dx} = - \cos x$.
Separating the variables,we get:
$\frac{dy}{1 + y} = - \frac{\cos x}{2 + \sin x} dx$.
Integrating both sides:
$\int \frac{dy}{1 + y} = - \int \frac{\cos x}{2 + \sin x} dx$.
Let $u = 2 + \sin x$,then $du = \cos x dx$.
So,$\ln|1 + y| = - \ln|2 + \sin x| + C$.
This can be written as $\ln|1 + y| + \ln|2 + \sin x| = C$,or $\ln|(1 + y)(2 + \sin x)| = C$.
Thus,$(1 + y)(2 + \sin x) = K$ (where $K = e^C$).
Given $y(0) = 1$,we substitute $x = 0$ and $y = 1$:
$(1 + 1)(2 + \sin 0) = K \implies 2(2 + 0) = K \implies K = 4$.
So,$(1 + y)(2 + \sin x) = 4$.
To find $y\left( \frac{\pi}{2} \right)$,substitute $x = \frac{\pi}{2}$:
$(1 + y)(2 + \sin \frac{\pi}{2}) = 4
(1 + y)(2 + 1) = 4
3(1 + y) = 4
1 + y = \frac{4}{3}
y = \frac{4}{3} - 1 = \frac{1}{3}$.

(B) Given the differential equation: ${e^{dy/dx}} = (x + 1)$.
Taking the natural logarithm on both sides: $\frac{dy}{dx} = \log(x + 1)$.
Separating the variables: $dy = \log(x + 1) dx$.
Integrating both sides: $y = \int \log(x + 1) dx$.
Using integration by parts,let $u = \log(x + 1)$ and $dv = dx$. Then $du = \frac{1}{x+1} dx$ and $v = x + 1$.
$y = (x + 1)\log(x + 1) - \int \frac{x+1}{x+1} dx$.
$y = (x + 1)\log(x + 1) - \int 1 dx$.
$y = (x + 1)\log(x + 1) - x + C$.
Given the initial condition $y(0) = 3$,substitute $x = 0$ and $y = 3$:
$3 = (0 + 1)\log(0 + 1) - 0 + C$.
$3 = 1 \cdot \log(1) + C$.
Since $\log(1) = 0$,we get $C = 3$.
Therefore,the solution is $y = (x + 1)\log |x + 1| - x + 3$.

(A) Given the differential equation: $\frac{dy}{dx} \tan y = \sin(x + y) + \sin(x - y)$.
Using the trigonometric identity $\sin(A + B) + \sin(A - B) = 2 \sin A \cos B$,we get:
$\frac{dy}{dx} \tan y = 2 \sin x \cos y$.
Rewrite $\tan y$ as $\frac{\sin y}{\cos y}$:
$\frac{dy}{dx} \cdot \frac{\sin y}{\cos y} = 2 \sin x \cos y$.
Separate the variables:
$\frac{\sin y}{\cos^2 y} dy = 2 \sin x dx$.
Integrate both sides:
$\int \frac{\sin y}{\cos^2 y} dy = 2 \int \sin x dx$.
Let $u = \cos y$,then $du = -\sin y dy$,so $\int -u^{-2} du = u^{-1} = \frac{1}{\cos y} = \sec y$.
Thus,$\sec y = -2 \cos x + c$,which simplifies to $\sec y + 2 \cos x = c$.

(B) Given the differential equation: $y\,dx + (x + x^2y)dy = 0$
Rearranging the terms,we get: $y\,dx + x\,dy + x^2y\,dy = 0$
$(y\,dx + x\,dy) = -x^2y\,dy$
Dividing both sides by $x^2y^2$ (assuming $x, y \neq 0$):
$\frac{y\,dx + x\,dy}{x^2y^2} = -\frac{dy}{y}$
Recognizing that $d(xy) = y\,dx + x\,dy$,the equation becomes:
$\frac{d(xy)}{(xy)^2} = -\frac{dy}{y}$
Integrating both sides:
$\int (xy)^{-2} d(xy) = -\int \frac{1}{y} dy$
$-\frac{1}{xy} = -\log|y| + c$
Rearranging the terms,we get:
$-\frac{1}{xy} + \log|y| = c$

(B) Given differential equation is $(x - y^3)dx + 3xy^2dy = 0$.
Rearranging the terms,we get $x dx - y^3 dx + 3xy^2 dy = 0$.
Let $t = y^3$,then $dt = 3y^2 dy$.
Substituting these into the equation,we get $x dx - t dx + x dt = 0$.
Dividing by $x^2$,we get $\frac{x dx - t dx + x dt}{x^2} = 0$,which simplifies to $\frac{dx}{x} + \frac{x dt - t dx}{x^2} = 0$.
This is equivalent to $d(\log x) + d(\frac{t}{x}) = 0$.
Integrating both sides,we get $\log x + \frac{t}{x} = k$.
Substituting $t = y^3$ back,the solution is $\log x + \frac{y^3}{x} = k$.

(B) Given differential equation is $x \, dy + y \, dx - \sqrt{1 - x^2 y^2} \, dx = 0$.
Rearranging the terms,we get $x \, dy + y \, dx = \sqrt{1 - x^2 y^2} \, dx$.
We know that $d(xy) = x \, dy + y \, dx$.
Substituting this into the equation,we get $d(xy) = \sqrt{1 - (xy)^2} \, dx$.
Dividing both sides by $\sqrt{1 - (xy)^2}$,we get $\frac{d(xy)}{\sqrt{1 - (xy)^2}} = dx$.
Integrating both sides,we get $\int \frac{d(xy)}{\sqrt{1 - (xy)^2}} = \int dx$.
This results in $\sin^{-1}(xy) = x + c$.
Taking the sine of both sides,we get $xy = \sin(x + c)$.

(A) Given the differential equation: $y \, dx - x \, dy + x y^2 \, dx = 0$.
Divide the entire equation by $y^2$ (assuming $y \neq 0$):
$\frac{y \, dx - x \, dy}{y^2} + x \, dx = 0$.
We recognize that $\frac{y \, dx - x \, dy}{y^2} = d\left( \frac{x}{y} \right)$.
Substituting this into the equation,we get:
$d\left( \frac{x}{y} \right) + x \, dx = 0$.
Integrating both sides:
$\int d\left( \frac{x}{y} \right) = - \int x \, dx$.
$\frac{x}{y} = - \frac{x^2}{2} + c$,where $c$ is the constant of integration.
Multiply by $2y$:
$2x = -x^2 y + 2cy$.
Rearranging the terms:
$2x + x^2 y = (2c)y$.
Let $\lambda = 2c$,then the solution is $2x + x^2 y = \lambda y$.

(D) Given the differential equation: $ydx - xdy = xy\,dx$
Divide both sides by $xy$ (assuming $x, y \neq 0$):
$\frac{ydx - xdy}{xy} = dx$
We recognize that $\frac{ydx - xdy}{x^2} = d(\frac{y}{x})$,but here we have $xy$ in the denominator. Let us rearrange:
$\frac{ydx - xdy}{xy} = dx$
$\frac{1}{x}dx - \frac{1}{y}dy = dx$
Integrating both sides:
$\int \frac{1}{x}dx - \int \frac{1}{y}dy = \int dx$
$\ln|x| - \ln|y| = x + C$
$\ln|\frac{x}{y}| = x + C$
Taking the exponential of both sides:
$\frac{x}{y} = e^{x+C} = e^C \cdot e^x$
Let $e^C = \frac{1}{c}$ (where $c$ is an arbitrary constant):
$\frac{x}{y} = \frac{1}{c} e^x$
$cx = y e^x$
Thus,the general solution is $y e^x = cx$.

(A) Given the slope of the curve is $\frac{dy}{dx} = \frac{2y}{x}$.
Separating the variables,we get $\frac{dy}{y} = \frac{2dx}{x}$.
Integrating both sides,we have $\int \frac{dy}{y} = 2 \int \frac{dx}{x}$.
This gives $\ln|y| = 2 \ln|x| + C$,which simplifies to $\ln|y| = \ln|x^2| + C$.
Taking the exponential of both sides,we get $y = c x^2$.
Since the curve passes through the point $(1, 1)$,we substitute $x = 1$ and $y = 1$ into the equation:
$1 = c(1)^2$,which implies $c = 1$.
Therefore,the equation of the curve is $y = x^2$.

(A) Given the differential equation: $\frac{dy}{dx} = \frac{-2xy}{x^2 + 1}$.
Separating the variables,we get: $\frac{dy}{y} = \frac{-2x}{x^2 + 1} dx$.
Integrating both sides: $\int \frac{dy}{y} = -\int \frac{2x}{x^2 + 1} dx$.
This gives: $\ln|y| = -\ln(x^2 + 1) + C$.
Using the property of logarithms: $\ln|y| + \ln(x^2 + 1) = C$,which simplifies to $\ln|y(x^2 + 1)| = C$.
Thus,$y(x^2 + 1) = e^C = k$ (where $k$ is a constant).
Since the curve passes through the point $(1, 2)$,substitute $x = 1$ and $y = 2$ into the equation:
$2(1^2 + 1) = k \implies 2(2) = k \implies k = 4$.
Therefore,the equation of the curve is $y(x^2 + 1) = 4$.

(B) Given the differential equation: $(1 + y^2)dx - xydy = 0$.
Rearranging the terms to separate the variables,we get: $\frac{dx}{x} = \frac{y dy}{1 + y^2}$.
Integrating both sides: $\int \frac{dx}{x} = \int \frac{y dy}{1 + y^2}$.
Let $u = 1 + y^2$,then $du = 2y dy$,so $y dy = \frac{du}{2}$.
Thus,$\ln|x| = \frac{1}{2} \ln(1 + y^2) + C$.
This can be written as $\ln|x| = \ln(\sqrt{1 + y^2}) + \ln c$,where $C = \ln c$.
So,$|x| = c\sqrt{1 + y^2}$.
Since the curve passes through $(1, 0)$,we substitute $x = 1$ and $y = 0$: $1 = c\sqrt{1 + 0^2} \implies c = 1$.
Therefore,the equation is $|x| = \sqrt{1 + y^2}$,which simplifies to $x^2 = 1 + y^2$ or $x^2 - y^2 = 1$.

(C) Given the differential equation $\frac{dy}{dx} = x + \frac{1}{x^2}$.
Integrating both sides with respect to $x$:
$\int dy = \int (x + x^{-2}) dx$
$y = \frac{x^2}{2} - \frac{1}{x} + c$
Since the curve passes through $(3, 9)$,we substitute $x = 3$ and $y = 9$ to find $c$:
$9 = \frac{3^2}{2} - \frac{1}{3} + c$
$9 = \frac{9}{2} - \frac{1}{3} + c$
$9 = \frac{27 - 2}{6} + c$
$9 = \frac{25}{6} + c$
$c = 9 - \frac{25}{6} = \frac{54 - 25}{6} = \frac{29}{6}$
Substituting $c$ back into the equation:
$y = \frac{x^2}{2} - \frac{1}{x} + \frac{29}{6}$
Multiply the entire equation by $6x$:
$6xy = 3x^3 - 6 + 29x$
$6xy = 3x^3 + 29x - 6$.

(A) Given the slope of the curve is $\frac{dy}{dx} = \frac{y - 1}{x^2 + x}$.
Separating the variables,we get $\frac{dy}{y - 1} = \frac{dx}{x(x + 1)}$.
Using partial fractions,$\frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{x + 1}$.
Integrating both sides,$\int \frac{dy}{y - 1} = \int (\frac{1}{x} - \frac{1}{x + 1}) dx$.
$\ln|y - 1| = \ln|x| - \ln|x + 1| + C$.
$\ln|y - 1| = \ln|\frac{x}{x + 1}| + C$.
$y - 1 = k \cdot \frac{x}{x + 1}$,where $k = e^C$.
Since the curve passes through $(1, 0)$,substitute $x = 1$ and $y = 0$:
$0 - 1 = k \cdot \frac{1}{1 + 1} \implies -1 = \frac{k}{2} \implies k = -2$.
Thus,$y - 1 = -2 \cdot \frac{x}{x + 1}$.
$(y - 1)(x + 1) = -2x$.
$(y - 1)(x + 1) + 2x = 0$.

(C) Given that the slope of the curve is the reciprocal of twice the ordinate $(y)$:
$\frac{dy}{dx} = \frac{1}{2y}$
By separating the variables,we get:
$2y \, dy = dx$
Integrating both sides:
$\int 2y \, dy = \int dx$
$y^2 = x + C$
Since the curve passes through the point $(4, 3)$,we substitute $x = 4$ and $y = 3$:
$3^2 = 4 + C$
$9 = 4 + C$
$C = 5$
Substituting the value of $C$ back into the equation,we get:
$y^2 = x + 5$

(B) Given the differential equation: $\frac{dx}{dt} = x + 1$.
Separating the variables,we get: $\frac{dx}{x + 1} = dt$.
Integrating both sides: $\int \frac{dx}{x + 1} = \int dt$.
This yields: $\ln(x + 1) = t + C$.
At $t = 0$,the distance $x = 0$. Substituting these values: $\ln(0 + 1) = 0 + C \implies \ln(1) = C \implies C = 0$.
Thus,the equation for time is: $t = \ln(x + 1)$.
To find the time taken to cover a distance of $x = 99 \ m$:
$t = \ln(99 + 1) = \ln(100)$.
Since $\ln(100) = \log_e(10^2) = 2 \log_e 10$.
Therefore,the correct option is $B$.

(B) Given differential equation is $x \, dy - y \, dx = 0$.
Rearranging the terms,we get $x \, dy = y \, dx$.
Separating the variables,we have $\frac{dy}{y} = \frac{dx}{x}$.
Integrating both sides,$\int \frac{dy}{y} = \int \frac{dx}{x}$.
This gives $\ln |y| = \ln |x| + \ln |c|$,where $\ln |c|$ is the constant of integration.
Using logarithmic properties,$\ln |y| = \ln |cx|$.
Taking the exponential of both sides,$y = cx$.
This is the equation of a straight line passing through the origin.

(C) Given the velocity of the particle is $\frac{dx}{dt} = \cos^2(\pi x)$.
To find the time $t$ taken to reach a position $x$,we separate the variables:
$dt = \frac{dx}{\cos^2(\pi x)} = \sec^2(\pi x) dx$.
Integrating both sides from the origin $(x=0, t=0)$ to a position $x$ at time $t$:
$\int_0^t dt = \int_0^x \sec^2(\pi u) du$.
$t = \left[ \frac{\tan(\pi u)}{\pi} \right]_0^x = \frac{\tan(\pi x)}{\pi}$.
As $x \to \frac{1}{2}$,$\tan(\pi x) \to \tan(\frac{\pi}{2}) \to \infty$.
Therefore,$t \to \infty$ as $x \to \frac{1}{2}$.
This implies that the particle takes an infinite amount of time to reach the point $x = \frac{1}{2}$,meaning it never reaches this point.

(B) Given the differential equation: $y - x\frac{dy}{dx} = a(y^2 + \frac{dy}{dx})$
Rearranging the terms to group $\frac{dy}{dx}$:
$y - ay^2 = x\frac{dy}{dx} + a\frac{dy}{dx}$
$y(1 - ay) = (x + a)\frac{dy}{dx}$
Separating the variables:
$\frac{dy}{y(1 - ay)} = \frac{dx}{x + a}$
Using partial fractions for the left side:
$\frac{1}{y(1 - ay)} = \frac{1}{y} + \frac{a}{1 - ay}$
Integrating both sides:
$\int (\frac{1}{y} + \frac{a}{1 - ay}) dy = \int \frac{1}{x + a} dx$
$\ln|y| - \ln|1 - ay| = \ln|x + a| + \ln|c|$
$\ln|\frac{y}{1 - ay}| = \ln|c(x + a)|$
Taking the exponential of both sides:
$\frac{y}{1 - ay} = c(x + a)$
$y = c(x + a)(1 - ay)$

(A) Given the differential equation $\sqrt{a + x} \frac{dy}{dx} + xy = 0$.
Rearranging the terms,we get $\frac{dy}{dx} = -\frac{xy}{\sqrt{a + x}}$.
Separating the variables,we have $\frac{dy}{y} = -\frac{x}{\sqrt{a + x}} dx$.
Integrating both sides,$\int \frac{dy}{y} = -\int \frac{x}{\sqrt{x + a}} dx$.
Let $u = x + a$,then $x = u - a$ and $dx = du$.
$\ln y = -\int \frac{u - a}{\sqrt{u}} du = -\int (u^{1/2} - a u^{-1/2}) du$.
$\ln y = -(\frac{2}{3} u^{3/2} - 2a u^{1/2}) + C$.
$\ln y = -\frac{2}{3} (x + a)^{3/2} + 2a(x + a)^{1/2} + \ln A$.
$\ln y = -\frac{2}{3} (x + a)^{1/2} (x + a) + 2a(x + a)^{1/2} + \ln A$.
$\ln y = \sqrt{x + a} [-\frac{2}{3}x - \frac{2}{3}a + 2a] + \ln A$.
$\ln y = \sqrt{x + a} [\frac{-2x - 2a + 6a}{3}] + \ln A = \frac{2}{3} \sqrt{x + a} (2a - x) + \ln A$.
Taking the exponential of both sides,$y = A e^{\frac{2}{3}(2a - x)\sqrt{x + a}}$.

(A) Given differential equation: ${x^4}\frac{{dy}}{{dx}} + {x^3}y + \text{cosec}(xy) = 0$
Multiply by $dx$: ${x^4}dy + {x^3}y\,dx + \text{cosec}(xy)\,dx = 0$
Rearrange the terms: ${x^3}(x\,dy + y\,dx) + \text{cosec}(xy)\,dx = 0$
Recognizing the differential of a product: ${x^3}d(xy) + \text{cosec}(xy)\,dx = 0$
Separate the variables: $\frac{d(xy)}{\text{cosec}(xy)} + \frac{dx}{x^3} = 0$
This simplifies to: $\sin(xy)\,d(xy) + x^{-3}\,dx = 0$
Integrating both sides: $\int \sin(xy)\,d(xy) + \int x^{-3}\,dx = \int 0$
Result of integration: $-\cos(xy) + \frac{x^{-2}}{-2} = C_1$
Multiply by $-2$: $2\cos(xy) + x^{-2} = c$ (where $c = -2C_1$).

(A) The slope of the curve is given by $\frac{dy}{dx} = \frac{y - 1}{x^2 + x}$.
Separating the variables,we get $\frac{dy}{y - 1} = \frac{dx}{x(x + 1)}$.
Using partial fractions,$\frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{x + 1}$.
Integrating both sides: $\int \frac{dy}{y - 1} = \int (\frac{1}{x} - \frac{1}{x + 1}) dx$.
$\ln|y - 1| = \ln|x| - \ln|x + 1| + C$.
$\ln|y - 1| = \ln|\frac{x}{x + 1}| + C$,which implies $y - 1 = k \cdot \frac{x}{x + 1}$.
Since the curve passes through $(1, 0)$,substitute $x = 1$ and $y = 0$: $0 - 1 = k \cdot \frac{1}{1 + 1} \Rightarrow -1 = \frac{k}{2} \Rightarrow k = -2$.
Thus,$y - 1 = -2 \cdot \frac{x}{x + 1} \Rightarrow (y - 1)(x + 1) = -2x$.
Therefore,$(y - 1)(x + 1) + 2x = 0$.

(C) Given that the slope $\frac{dy}{dx}$ is inversely proportional to $2y$,we have:
$\frac{dy}{dx} = \frac{1}{2y}$
Separating the variables,we get:
$2y \, dy = dx$
Integrating both sides:
$\int 2y \, dy = \int dx$
$y^2 = x + C$
Since the curve passes through $(4, 3)$,we substitute $x = 4$ and $y = 3$:
$(3)^2 = 4 + C$
$9 = 4 + C$
$C = 5$
Substituting the value of $C$ back into the equation,we get:
$y^2 = x + 5$

(D) Given the differential equation $\frac{dy}{dx} = y + 3$.
Separating the variables,we get $\frac{dy}{y + 3} = dx$.
Integrating both sides,we obtain $\int \frac{dy}{y + 3} = \int dx$,which gives $\ln|y + 3| = x + C$.
Since $y + 3 > 0$,we have $y + 3 = e^{x + C} = e^C \cdot e^x$.
Let $e^C = A$,so $y + 3 = A e^x$.
Using the initial condition $y(0) = 2$,we substitute $x = 0$ and $y = 2$:
$2 + 3 = A e^0 \Rightarrow A = 5$.
Thus,the particular solution is $y + 3 = 5 e^x$,or $y = 5 e^x - 3$.
To find $y(\ln 2)$,substitute $x = \ln 2$:
$y(\ln 2) = 5 e^{\ln 2} - 3$.
Since $e^{\ln 2} = 2$,we have $y(\ln 2) = 5(2) - 3 = 10 - 3 = 7$.

(NONE) Given the differential equation: $y(1 + xy)dx = xdy$.
Rearranging the terms: $ydx + xy^2 dx = xdy$.
$xy^2 dx = xdy - ydx$.
Dividing both sides by $xy^2$ (assuming $x, y \neq 0$): $dx = \frac{xdy - ydx}{xy^2} = \frac{1}{x} \cdot \frac{xdy - ydx}{y^2}$.
This can be written as: $dx = \frac{1}{x} d(\frac{x}{y})$.
Integrating both sides: $\int dx = \int \frac{1}{x} d(\frac{x}{y})$ is not the direct path. Let us rewrite: $\frac{xdy - ydx}{y^2} = x dx$.
This is $d(\frac{x}{y}) = x dx$.
Integrating both sides: $\frac{x}{y} = \frac{x^2}{2} + C$.
Since the curve passes through $(1, -1)$,substitute $x=1, y=-1$: $\frac{1}{-1} = \frac{1^2}{2} + C$.
$-1 = \frac{1}{2} + C \Rightarrow C = -\frac{3}{2}$.
So,$\frac{x}{y} = \frac{x^2 - 3}{2} \Rightarrow y = \frac{2x}{x^2 - 3}$.
Now,find $f(-\frac{1}{2})$: $f(-\frac{1}{2}) = \frac{2(-\frac{1}{2})}{(-\frac{1}{2})^2 - 3} = \frac{-1}{\frac{1}{4} - 3} = \frac{-1}{-\frac{11}{4}} = \frac{4}{11}$.

(B) Given the differential equation: $(2 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$.
This can be rewritten as: $\frac{d}{dx} [(2 + \sin x)(y + 1)] = 0$.
Integrating both sides with respect to $x$,we get: $(2 + \sin x)(y + 1) = C$,where $C$ is a constant.
Using the initial condition $y(0) = 1$,we substitute $x = 0$ and $y = 1$:
$(2 + \sin 0)(1 + 1) = C \Rightarrow (2 + 0)(2) = C \Rightarrow C = 4$.
Thus,the equation becomes: $(2 + \sin x)(y + 1) = 4$.
Solving for $y$: $y + 1 = \frac{4}{2 + \sin x} \Rightarrow y = \frac{4}{2 + \sin x} - 1$.
Now,we find $y(\frac{\pi}{2})$:
$y(\frac{\pi}{2}) = \frac{4}{2 + \sin(\frac{\pi}{2})} - 1 = \frac{4}{2 + 1} - 1 = \frac{4}{3} - 1 = \frac{1}{3}$.

(C) Given the differential equation: $\sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0$
Divide both sides by $\tan x \tan y$:
$\frac{\sec^2 x}{\tan x} \, dx + \frac{\sec^2 y}{\tan y} \, dy = 0$
Integrating both sides:
$\int \frac{\sec^2 x}{\tan x} \, dx + \int \frac{\sec^2 y}{\tan y} \, dy = \int 0 \, dx$
Using the substitution $u = \tan x$ $(du = \sec^2 x \, dx)$ and $v = \tan y$ $(dv = \sec^2 y \, dy)$,we get:
$\ln|\tan x| + \ln|\tan y| = \ln|c|$
Using the property $\ln A + \ln B = \ln(AB)$:
$\ln|\tan x \tan y| = \ln|c|$
$\tan x \tan y = c$
Since $\tan y = \frac{1}{\cot y}$,we can rewrite this as:
$\tan x = c \cot y$.

(B) Given the differential equation $\frac{dy}{dx} = 1 + x + y + xy$.
Factorizing the right side: $\frac{dy}{dx} = (1 + x) + y(1 + x) = (1 + x)(1 + y)$.
Separating the variables: $\frac{dy}{1 + y} = (1 + x) dx$.
Integrating both sides: $\int \frac{dy}{1 + y} = \int (1 + x) dx$.
This gives $\log_e(1 + y) = x + \frac{x^2}{2} + C$.
Exponentiating both sides: $1 + y = e^{x + x^2/2 + C} = k e^{x + x^2/2}$,where $k = e^C$.
So,$y = k e^{x + x^2/2} - 1$.
Using the initial condition $y(-1) = 0$: $0 = k e^{-1 + 1/2} - 1 \implies 0 = k e^{-1/2} - 1 \implies k = e^{1/2}$.
Substituting $k$ back: $y = e^{1/2} e^{x + x^2/2} - 1 = e^{(1 + 2x + x^2)/2} - 1 = e^{(1 + x)^2/2} - 1$.

(D) Given the differential equation $\frac{dy}{dx} = 1 + x + y^2 + xy^2$.
Factoring the right side,we get $\frac{dy}{dx} = (1 + x) + y^2(1 + x) = (1 + x)(1 + y^2)$.
Separating the variables,we have $\frac{dy}{1 + y^2} = (1 + x) dx$.
Integrating both sides,$\int \frac{dy}{1 + y^2} = \int (1 + x) dx$.
This yields $\tan^{-1} y = x + \frac{x^2}{2} + c$.
Using the initial condition $y(0) = 0$,we substitute $x = 0$ and $y = 0$: $\tan^{-1}(0) = 0 + \frac{0^2}{2} + c$,which gives $c = 0$.
Therefore,$\tan^{-1} y = x + \frac{x^2}{2}$,which implies $y = \tan \left( x + \frac{x^2}{2} \right)$.

(C) Let the point on the curve be $(x, y)$.
The slope of the tangent at $(x, y)$ is given by $\frac{dy}{dx}$.
The slope of the line joining $(x, y)$ to the origin $(0, 0)$ is $\frac{y - 0}{x - 0} = \frac{y}{x}$.
According to the given condition,$\frac{dy}{dx} = 2 \left( \frac{y}{x} \right)$.
This is a variable separable differential equation: $\frac{dy}{y} = 2 \frac{dx}{x}$.
Integrating both sides,we get $\int \frac{dy}{y} = 2 \int \frac{dx}{x}$.
$\ln|y| = 2 \ln|x| + C$,which can be written as $\ln|y| = \ln|x^2| + \ln|c|$.
Thus,$y = cx^2$,which represents a parabola.