Variable separable type differential equations Questions in English

(D) The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = xy$.
Separating the variables,we get $\frac{1}{y} dy = x dx$.
Integrating both sides,we have $\int \frac{1}{y} dy = \int x dx$,which gives $\log y = \frac{x^{2}}{2} + C$.
Since the curve passes through the point $(1, 1)$,we substitute $x = 1$ and $y = 1$ into the equation:
$\log(1) = \frac{1^{2}}{2} + C \implies 0 = \frac{1}{2} + C \implies C = -\frac{1}{2}$.
Substituting the value of $C$ back into the equation,we get $\log y = \frac{x^{2}}{2} - \frac{1}{2}$.
Multiplying by $2$,we obtain $2 \log y = x^{2} - 1$.

(D) Let the equation of the curve be $y=f(x)$.
Since the curve passes through the point $(1,1)$,we have $f(1)=1$.
According to the problem,at any point $(x, y)$ on the curve,the product of the slope of its tangent $(\frac{dy}{dx})$ and the $x$-coordinate is equal to the $y$-coordinate.
Thus,$x \cdot \frac{dy}{dx} = y$.
Rearranging the terms,we get $\frac{dy}{y} = \frac{dx}{x}$.
Integrating both sides,we obtain $\int \frac{dy}{y} = \int \frac{dx}{x}$,which gives $\ln|y| = \ln|x| + C$.
This simplifies to $y = kx$,where $k = e^C$.
Using the condition $f(1)=1$,we substitute $x=1$ and $y=1$ into $y=kx$ to get $1 = k(1)$,so $k=1$.
The equation of the curve is $y=x$.
Checking the given options,the point $(2,2)$ satisfies the equation $y=x$.

(D) Given the differential equation $e^{dy/dx} = x+1$.
Taking the natural logarithm on both sides,we get $\frac{dy}{dx} = \log(x+1)$.
Integrating both sides with respect to $x$,we have $\int dy = \int \log(x+1) dx$.
Using integration by parts,$\int \log(x+1) dx = (x+1) \log(x+1) - (x+1) + C$.
Alternatively,$y = x \log(x+1) - \int \frac{x}{x+1} dx = x \log(x+1) - \int (1 - \frac{1}{x+1}) dx = x \log(x+1) - x + \log(x+1) + C$.
Thus,$y = (x+1) \log(x+1) - x + C$.
Given the initial condition $y(0) = 3$,we substitute $x=0$ and $y=3$: $3 = (0+1) \log(1) - 0 + C \Rightarrow 3 = 0 - 0 + C \Rightarrow C = 3$.
Substituting $C=3$ into the general solution,we get $y = (x+1) \log(x+1) - x + 3$.
Rearranging the terms,we obtain $y+x-3 = (x+1) \log(x+1)$.

(A) Given the differential equation: $\frac{dy}{dx} = (x+y)^2$ $(i)$
Let $x+y = t$.
Differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dt}{dx}$,which implies $\frac{dy}{dx} = \frac{dt}{dx} - 1$.
Substituting this into equation $(i)$:
$\frac{dt}{dx} - 1 = t^2$
$\frac{dt}{dx} = t^2 + 1$
Separating the variables:
$\frac{dt}{t^2 + 1} = dx$
Integrating both sides:
$\int \frac{dt}{t^2 + 1} = \int dx$
$\tan^{-1}(t) = x + C$
Substituting $t = x+y$ back:
$\tan^{-1}(x+y) = x + C$

(C) Given differential equation is,$\frac{dy}{y} + \frac{dx}{x} = 0$.
Integrating both sides with respect to their variables:
$\int \frac{dy}{y} + \int \frac{dx}{x} = \int 0 \, dx$
Using the standard integral formula $\int \frac{1}{u} du = \log |u| + C$,we get:
$\log |y| + \log |x| = \log |c|$
Using the logarithmic property $\log a + \log b = \log(ab)$:
$\log |xy| = \log |c|$
Taking the exponential on both sides:
$xy = c$

(D) Given the differential equation $e^{dy/dx} = x$.
Taking the natural logarithm on both sides,we get:
$\frac{dy}{dx} = \log x$
Integrating both sides with respect to $x$:
$\int dy = \int \log x \, dx$
Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = \log x$ and $dv = dx$:
$y = x \log x - \int x \cdot \frac{1}{x} \, dx$
$y = x \log x - \int 1 \, dx$
$y = x \log x - x + C$
$y = x(\log x - 1) + C$ (Equation $i$)
Given the initial conditions $x = 1$ and $y = 0$,substitute these into Equation $i$:
$0 = 1(\log 1 - 1) + C$
$0 = 1(0 - 1) + C$
$0 = -1 + C$
$C = 1$
Substituting $C = 1$ back into Equation $i$,we get:
$y = x(\log x - 1) + 1$

(A) Given the differential equation: $\left(\frac{dy}{dx}\right)^{2} = 1 - x^{2} - y^{2} + x^{2}y^{2}$
Factor the right side: $\left(\frac{dy}{dx}\right)^{2} = (1 - x^{2}) - y^{2}(1 - x^{2}) = (1 - x^{2})(1 - y^{2})$
Taking the square root on both sides: $\frac{dy}{dx} = \sqrt{1 - x^{2}} \sqrt{1 - y^{2}}$
Separate the variables: $\frac{dy}{\sqrt{1 - y^{2}}} = \sqrt{1 - x^{2}} dx$
Integrate both sides: $\int \frac{dy}{\sqrt{1 - y^{2}}} = \int \sqrt{1 - x^{2}} dx$
Using standard integrals $\int \frac{1}{\sqrt{1 - y^{2}}} dy = \sin^{-1} y$ and $\int \sqrt{1 - x^{2}} dx = \frac{x}{2} \sqrt{1 - x^{2}} + \frac{1}{2} \sin^{-1} x + C_1$:
$\sin^{-1} y = \frac{x}{2} \sqrt{1 - x^{2}} + \frac{1}{2} \sin^{-1} x + C_1$
Multiply by $2$: $2 \sin^{-1} y = x \sqrt{1 - x^{2}} + \sin^{-1} x + 2C_1$
Let $C = 2C_1$,then: $2 \sin^{-1} y = x \sqrt{1 - x^{2}} + \sin^{-1} x + C$

(C) Given the differential equation: $x dy - y dx = 0$
Rearranging the terms,we get: $x dy = y dx$
Dividing both sides by $xy$ (assuming $x, y \neq 0$): $\frac{dy}{y} = \frac{dx}{x}$
Integrating both sides: $\int \frac{1}{y} dy = \int \frac{1}{x} dx$
This gives: $\ln|y| = \ln|x| + C$
Taking the exponential of both sides: $y = cx$,where $c$ is an arbitrary constant.
This is the equation of a straight line passing through the origin.

(A) Given differential equation is $x \frac{dy}{dx} - y = 3$.
Rearranging the terms,we get $x \frac{dy}{dx} = y + 3$.
Separating the variables,we have $\frac{dy}{y + 3} = \frac{dx}{x}$.
Integrating both sides,we get $\int \frac{dy}{y + 3} = \int \frac{dx}{x}$.
This results in $\ln|y + 3| = \ln|x| + \ln|c|$,where $c$ is the constant of integration.
Taking the exponential of both sides,we get $y + 3 = cx$,or $y = cx - 3$.
This is the equation of a family of straight lines passing through the point $(0, -3)$.

(B) The given differential equation is $2x \frac{dy}{dx} - y = 3$.
Rearranging the terms,we get $2x \frac{dy}{dx} = y + 3$.
Separating the variables,we have $\frac{dy}{y+3} = \frac{dx}{2x}$.
Integrating both sides,we get $\int \frac{dy}{y+3} = \frac{1}{2} \int \frac{dx}{x}$.
This gives $\ln|y+3| = \frac{1}{2} \ln|x| + C_1$.
Multiplying by $2$,we get $2 \ln|y+3| = \ln|x| + 2C_1$.
Using properties of logarithms,$\ln(y+3)^2 = \ln|x| + \ln|c|$,where $c = e^{2C_1}$.
Thus,$(y+3)^2 = cx$,which is the equation of a family of parabolas.

(C) Given differential equation is $\sqrt{1-x^{2} y^{2}} \cdot dx = y \cdot dx + x \cdot dy$.
We know that $d(xy) = y \cdot dx + x \cdot dy$.
Substituting this into the equation,we get $\sqrt{1-(xy)^{2}} \cdot dx = d(xy)$.
Rearranging the terms,we have $dx = \frac{d(xy)}{\sqrt{1-(xy)^{2}}}$.
Integrating both sides,we get $\int dx = \int \frac{d(xy)}{\sqrt{1-(xy)^{2}}}$.
This results in $x = \sin^{-1}(xy) + C$.
Taking the sine of both sides,we get $\sin(x - C) = xy$,which is equivalent to $\sin(x + C) = xy$ (where $C$ is an arbitrary constant).

(D) Given the slope of the tangent $\frac{dy}{dx} = \frac{3x}{y}$.
By separating the variables,we get $y \, dy = 3x \, dx$.
Integrating both sides,we have $\int y \, dy = \int 3x \, dx$,which gives $\frac{y^2}{2} = \frac{3x^2}{2} + C$.
Multiplying by $2$,we get $y^2 = 3x^2 + 2C$,or $y^2 - 3x^2 = K$ where $K = 2C$.
Since the curve passes through $(1, 2)$,we substitute $x=1$ and $y=2$ into the equation: $2^2 - 3(1)^2 = K$,so $4 - 3 = K$,which means $K = 1$.
The equation of the curve is $y^2 - 3x^2 = 1$.
This equation is of the form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$,which represents a Hyperbola.

(B) Given $f^{\prime}(x) = \sqrt{f^2(x)-1}$.
We can write this as $\frac{f^{\prime}(x)}{\sqrt{f^2(x)-1}} = 1$.
Integrating both sides with respect to $x$,we get $\int \frac{f^{\prime}(x)}{\sqrt{f^2(x)-1}} dx = \int 1 dx$.
Using the standard integral $\int \frac{1}{\sqrt{t^2-1}} dt = \log |t + \sqrt{t^2-1}| + C$,we have $\log |f(x) + \sqrt{f^2(x)-1}| = x + C$.
Given $f(0) = 1$,substituting $x=0$ gives $\log |f(0) + \sqrt{f^2(0)-1}| = 0 + C$.
$\log |1 + \sqrt{1-1}| = C \Rightarrow \log(1) = C \Rightarrow C = 0$.
So,$\log |f(x) + \sqrt{f^2(x)-1}| = x$.
At $x=1$,$\log |f(1) + \sqrt{f^2(1)-1}| = 1$.
Taking the exponential of both sides,$f(1) + \sqrt{f^2(1)-1} = e^1 = e$.
$\sqrt{f^2(1)-1} = e - f(1)$.
Squaring both sides,$f^2(1) - 1 = e^2 + f^2(1) - 2ef(1)$.
$-1 = e^2 - 2ef(1) \Rightarrow 2ef(1) = e^2 + 1$.
Thus,$f(1) = \frac{e^2+1}{2e}$.

(C) Given the slope of the tangent $\frac{dy}{dx} = \frac{y-4}{x-3}$.
Separating the variables,we get $\int \frac{dy}{y-4} = \int \frac{dx}{x-3}$.
Integrating both sides,$\ln|y-4| = \ln|x-3| + C$.
Since the curve passes through $(4, 3)$,we substitute $x=4$ and $y=3$: $\ln|3-4| = \ln|4-3| + C \Rightarrow \ln(1) = \ln(1) + C \Rightarrow C = 0$.
Thus,$\ln|y-4| = \ln|x-3|$,which implies $|y-4| = |x-3|$.
This gives two cases: $y-4 = x-3$ or $y-4 = -(x-3)$.
Case $1$: $y-x = 1$. This line does not intersect $y=x$ except at infinity.
Case $2$: $y-4 = -x+3 \Rightarrow x+y = 7$.
To find the intersection with $y=x$,substitute $y=x$ into $x+y=7$: $x+x=7 \Rightarrow 2x=7 \Rightarrow x=\frac{7}{2}$.
Thus,$y=\frac{7}{2}$. The point is $(\frac{7}{2}, \frac{7}{2})$.

(D) Given the differential equation: $y y^{\prime} = x \left[ \frac{y^2}{x^2} + \frac{\phi\left(\frac{y^2}{x^2}\right)}{\phi^{\prime}\left(\frac{y^2}{x^2}\right)} \right]$.
Divide both sides by $x^2$: $\frac{y}{x} \frac{dy}{dx} = \frac{y^2}{x^2} + \frac{\phi\left(\frac{y^2}{x^2}\right)}{\phi^{\prime}\left(\frac{y^2}{x^2}\right)}$.
Let $v = \frac{y^2}{x^2}$. Then $v x^2 = y^2$. Differentiating with respect to $x$: $2v x + x^2 \frac{dv}{dx} = 2y \frac{dy}{dx}$.
Thus,$y \frac{dy}{dx} = v x + \frac{x^2}{2} \frac{dv}{dx}$.
Substituting into the equation: $\frac{1}{x} (v x + \frac{x^2}{2} \frac{dv}{dx}) = v + \frac{\phi(v)}{\phi^{\prime}(v)}$.
$v + \frac{x}{2} \frac{dv}{dx} = v + \frac{\phi(v)}{\phi^{\prime}(v)}$.
$\frac{x}{2} \frac{dv}{dx} = \frac{\phi(v)}{\phi^{\prime}(v)}$.
Separating variables: $\frac{\phi^{\prime}(v)}{\phi(v)} dv = \frac{2}{x} dx$.
Integrating both sides: $\ln|\phi(v)| = 2 \ln|x| + \ln|c| = \ln|c x^2|$.
Therefore,$\phi(v) = c x^2$.
Substituting $v = \frac{y^2}{x^2}$,we get $\phi\left(\frac{y^2}{x^2}\right) = c x^2$.

(A) Given the differential equation: $\sec(x-y+1) dy = dx$.
Let $v = x-y+1$.
Then,$\frac{dv}{dx} = 1 - \frac{dy}{dx}$,which implies $\frac{dy}{dx} = 1 - \frac{dv}{dx}$.
The equation can be rewritten as $\frac{dy}{dx} = \cos(x-y+1)$.
Substituting $v$,we get $1 - \frac{dv}{dx} = \cos(v)$.
Rearranging the terms: $\frac{dv}{dx} = 1 - \cos(v)$.
Separating the variables: $\frac{dv}{1 - \cos(v)} = dx$.
Using the identity $1 - \cos(v) = 2\sin^2(\frac{v}{2})$,we have $\frac{dv}{2\sin^2(\frac{v}{2})} = dx$.
This simplifies to $\frac{1}{2} \csc^2(\frac{v}{2}) dv = dx$.
Integrating both sides: $\int \frac{1}{2} \csc^2(\frac{v}{2}) dv = \int dx$.
$-\cot(\frac{v}{2}) = x + c$.
Substituting $v = x-y+1$ back: $-\cot(\frac{x-y+1}{2}) = x + c$,which can be written as $x + \cot(\frac{x-y+1}{2}) = c'$ (where $c' = -c$).
Thus,the correct option is $A$.

(C) Given the differential equation: $x^2(y+1) \frac{dy}{dx} + y^2(x+1)^2 = 0$.
Rearranging the terms to separate the variables: $\frac{y+1}{y^2} dy = -\frac{(x+1)^2}{x^2} dx$.
Integrating both sides: $\int (\frac{1}{y} + \frac{1}{y^2}) dy = -\int (\frac{x^2 + 2x + 1}{x^2}) dx$.
$\int (\frac{1}{y} + y^{-2}) dy = -\int (1 + \frac{2}{x} + \frac{1}{x^2}) dx$.
$\log |y| - \frac{1}{y} = -(x + 2 \log |x| - \frac{1}{x}) + C$.
$\log |y| - \frac{1}{y} = -x - 2 \log |x| + \frac{1}{x} + C$.
$\log |y| + 2 \log |x| = \frac{1}{x} + \frac{1}{y} - x + C$.
$\log |x^2 y| = \frac{1}{x} + \frac{1}{y} - x + C$.
Given $y(1) = 2$,substitute $x=1$ and $y=2$: $\log |1^2 \times 2| = \frac{1}{1} + \frac{1}{2} - 1 + C$.
$\log 2 = 1 + 0.5 - 1 + C \implies \log 2 = 0.5 + C \implies C = \log 2 - 0.5$.
Substituting $C$ back: $\log |x^2 y| = \frac{1}{x} + \frac{1}{y} - x + \log 2 - 0.5$.
$\log |x^2 y| - \log 2 = \frac{1}{x} + \frac{1}{y} - x - 0.5$.
$\log |\frac{x^2 y}{2}| = \frac{1}{x} + \frac{1}{y} - x - 0.5$.
This matches option $C$.

(A) Given differential equation is $xy(y+2) dy + (y^3-1) dx = 0$.
Rearranging the terms,we get $\frac{dx}{x} + \frac{y(y+2)}{y^3-1} dy = 0$.
Integrating both sides,$\int \frac{dx}{x} + \int \frac{y^2+2y}{y^3-1} dy = c$.
Using partial fractions for $\frac{y^2+2y}{(y-1)(y^2+y+1)} = \frac{A}{y-1} + \frac{By+C}{y^2+y+1}$.
Solving for constants,we find $A = 1$,$B = 0$,$C = 1$.
So,$\int \frac{dx}{x} + \int \frac{1}{y-1} dy + \int \frac{1}{y^2+y+1} dy = c$.
$\log |x| + \log |y-1| + \int \frac{1}{(y+1/2)^2 + (\sqrt{3}/2)^2} dy = c$.
$\log |x(y-1)| + \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2y+1}{\sqrt{3}} \right) = c$.
Since $y^3-1 = (y-1)(y^2+y+1)$,the solution is $\log |x| + \frac{1}{3} \log |y^3-1| + \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2y+1}{\sqrt{3}} \right) = c$.

(B) Given the differential equation: $\cos(x+y) dy = dx$.
Let $v = x+y$. Then,differentiating with respect to $x$,we get $\frac{dv}{dx} = 1 + \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$.
The equation can be rewritten as $\frac{dy}{dx} = \frac{1}{\cos(x+y)} = \sec(x+y)$.
Substituting $v$ and $\frac{dy}{dx}$: $\frac{dv}{dx} - 1 = \sec(v)$.
$\frac{dv}{dx} = 1 + \sec(v) = 1 + \frac{1}{\cos(v)} = \frac{\cos(v)+1}{\cos(v)}$.
Separating variables: $\frac{\cos(v)}{\cos(v)+1} dv = dx$.
Using the identity $\cos(v) = 2\cos^2(v/2) - 1$,we have $\frac{2\cos^2(v/2)-1}{2\cos^2(v/2)} dv = dx$.
$(1 - \frac{1}{2}\sec^2(v/2)) dv = dx$.
Integrating both sides: $\int (1 - \frac{1}{2}\sec^2(v/2)) dv = \int dx$.
$v - \tan(v/2) = x + c$.
Substituting $v = x+y$: $(x+y) - \tan(\frac{x+y}{2}) = x + c$.
$y - \tan(\frac{x+y}{2}) = c$,or $y = \tan(\frac{x+y}{2}) + c$.

(C) Given the differential equation: $x^3 \sin y \frac{d y}{d x}=2$.
Separating the variables,we get: $\sin y \, dy = \frac{2}{x^3} \, dx$.
Integrating both sides: $\int \sin y \, dy = \int 2x^{-3} \, dx$.
This yields: $-\cos y = 2 \cdot \frac{x^{-2}}{-2} + C = -\frac{1}{x^2} + C$.
Multiplying by $-1$,we get: $\cos y = \frac{1}{x^2} - C$ ... $(i)$.
Given the condition $\lim _{x \rightarrow \infty} y(x) = \frac{\pi}{2}$,we take the limit as $x \rightarrow \infty$ on both sides of equation $(i)$:
$\lim _{x \rightarrow \infty} \cos y = \lim _{x \rightarrow \infty} \left( \frac{1}{x^2} - C \right)$.
Since $\cos$ is a continuous function,$\cos(\lim _{x \rightarrow \infty} y) = 0 - C$.
$\cos(\frac{\pi}{2}) = -C \Rightarrow 0 = -C \Rightarrow C = 0$.
Substituting $C = 0$ into equation $(i)$,we get: $\cos y = \frac{1}{x^2}$.

(A) The slope of the tangent to the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = 1 - \frac{1}{x^2}$.
Integrating both sides with respect to $x$:
$\int dy = \int \left( 1 - \frac{1}{x^2} \right) dx$
$y = x - (-\frac{1}{x}) + C$
$y = x + \frac{1}{x} + C$
Since the curve passes through the point $(1, 2)$,we substitute $x = 1$ and $y = 2$ to find the constant $C$:
$2 = 1 + \frac{1}{1} + C$
$2 = 1 + 1 + C$
$2 = 2 + C$
$C = 0$
Therefore,the equation of the curve is $y = x + \frac{1}{x}$.

(D) Given differential equation: $(1+y^2) dx - xy dy = 0$
Rearranging the terms to separate the variables: $\frac{dx}{x} = \frac{y}{1+y^2} dy$
Integrating both sides: $\int \frac{dx}{x} = \int \frac{y}{1+y^2} dy$
$\ln |x| = \frac{1}{2} \ln(1+y^2) + C$
Multiplying by $2$: $2 \ln |x| = \ln(1+y^2) + 2C$
$\ln(x^2) - \ln(1+y^2) = C'$ (where $C' = 2C$)
$\ln \left( \frac{x^2}{1+y^2} \right) = C'$
$\frac{x^2}{1+y^2} = e^{C'} = k$
Given $y(1) = 0$,substitute $x=1$ and $y=0$: $\frac{1^2}{1+0^2} = k \Rightarrow k = 1$
So,$\frac{x^2}{1+y^2} = 1 \Rightarrow x^2 = 1+y^2 \Rightarrow x^2 - y^2 = 1$
This equation represents a hyperbola.

(A) Given the differential equation: $\frac{dy}{dx} = \cos^2(x-y-1)$ $(i)$
Let $x-y-1 = p$.
Differentiating with respect to $x$,we get $1 - \frac{dy}{dx} = \frac{dp}{dx}$,which implies $\frac{dy}{dx} = 1 - \frac{dp}{dx}$.
Substituting this into equation $(i)$:
$1 - \frac{dp}{dx} = \cos^2 p$
$\frac{dp}{dx} = 1 - \cos^2 p = \sin^2 p$
$\frac{dp}{\sin^2 p} = dx$
$\operatorname{cosec}^2 p \, dp = dx$
Integrating both sides:
$\int \operatorname{cosec}^2 p \, dp = \int dx$
$-\cot p = x + C'$
$x = -C' - \cot p$
Let $C = -C'$,then $x = C - \cot(x-y-1)$.

(A) Given the differential equation: $\frac{dx}{dy} = \frac{\sin y(1 + y \cot y)}{x \log(x^2 e)}$.
Rearranging the terms,we get: $x \log(x^2 e) dx = (\sin y + y \cos y) dy$.
Integrating both sides: $\int x \log(x^2 e) dx = \int (\sin y + y \cos y) dy$.
For the left side,let $t = x^2 e$,then $dt = 2x e dx$,so $x dx = \frac{dt}{2e}$.
$\int \log(t) \frac{dt}{2e} = \frac{1}{2e} (t \log t - t) = \frac{x^2 e}{2e} (\log(x^2 e) - 1) = \frac{x^2}{2} (\log x^2 + \log e - 1) = \frac{x^2}{2} (2 \log x + 1 - 1) = x^2 \log x$.
For the right side,using integration by parts: $\int (\sin y + y \cos y) dy = y \sin y - \int \sin y dy + \int \sin y dy = y \sin y + C$.
Thus,$x^2 \log x = y \sin y + C$.
Given $y(1) = 0$,substitute $x = 1$ and $y = 0$: $1^2 \log(1) = 0 \cdot \sin(0) + C \implies 0 = 0 + C \implies C = 0$.
Therefore,the particular solution is $x^2 \log x = y \sin y$.

(A) Given the differential equation: $\frac{dy}{dx} = \cos^2(3x+y)$.
Let $3x+y = t$. Then,$3 + \frac{dy}{dx} = \frac{dt}{dx}$,so $\frac{dy}{dx} = \frac{dt}{dx} - 3$.
Substituting this into the equation: $\frac{dt}{dx} - 3 = \cos^2 t$,which gives $\frac{dt}{dx} = \cos^2 t + 3$.
Separating variables: $\frac{dt}{\cos^2 t + 3} = dx$.
Integrating both sides: $\int \frac{dt}{\cos^2 t + 3} = \int dx$.
Multiply numerator and denominator by $\sec^2 t$: $\int \frac{\sec^2 t dt}{1 + 3\sec^2 t} = \int dx$.
Using $\sec^2 t = 1 + \tan^2 t$: $\int \frac{\sec^2 t dt}{1 + 3(1 + \tan^2 t)} = \int \frac{\sec^2 t dt}{4 + 3\tan^2 t} = \int dx$.
Let $\tan t = m$,then $\sec^2 t dt = dm$.
The integral becomes $\int \frac{dm}{4 + 3m^2} = \frac{1}{3} \int \frac{dm}{\frac{4}{3} + m^2} = \frac{1}{3} \cdot \frac{1}{\frac{2}{\sqrt{3}}} \tan^{-1}\left(\frac{m}{2/\sqrt{3}}\right) = x + C$.
Simplifying: $\frac{1}{2\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3}m}{2}\right) = x + C$.
Multiplying by $2\sqrt{3}$: $\tan^{-1}\left(\frac{\sqrt{3}}{2} \tan t\right) = 2\sqrt{3}(x+C)$.
Since $t = 3x+y$,we have $\tan^{-1}\left(\frac{\sqrt{3}}{2} \tan(3x+y)\right) = 2\sqrt{3}(x+C)$.
Thus,$f(x) = 2\sqrt{3}(x+C)$.

(A) The given differential equation is $\frac{dy}{dx} = \frac{ax+b}{cy+d}$.
By separating the variables,we get $(cy+d)dy = (ax+b)dx$.
Integrating both sides,we have $\int (cy+d)dy = \int (ax+b)dx$.
This results in $\frac{cy^2}{2} + dy = \frac{ax^2}{2} + bx + K$,where $K$ is the constant of integration.
For the solution to represent a family of straight lines,the terms involving $x^2$ and $y^2$ must vanish.
This implies that the coefficients of $x^2$ and $y^2$ must be zero,so $a=0$ and $c=0$.
Substituting $a=0$ and $c=0$ into the equation,we get $dy = bdx$,which represents a family of lines provided that $b$ and $d$ are not both zero,i.e.,$b^2+d^2 \neq 0$.

(B) Given the differential equation $\frac{dy}{dx}=\frac{ax+by+c}{a_1x+b_1y+c_1}$.
If $\frac{a}{a_1}=\frac{b}{b_1}=m$,then the equation can be written as $\frac{dy}{dx}=\frac{m(a_1x+b_1y)+c}{a_1x+b_1y+c_1}$.
To solve this,we substitute $z = a_1x + b_1y$.
This substitution allows us to express the equation in a form where the variables $z$ and $x$ can be separated.

(D) Given the differential equation $\frac{dy}{dx} = \frac{y^3 \cos \sqrt{x}}{\sqrt{x} e^{1/y^2}}$.
Separating the variables,we get $\int \frac{e^{1/y^2}}{y^3} dy = \int \frac{\cos \sqrt{x}}{\sqrt{x}} dx$.
Let $t = \frac{1}{y^2}$,then $dt = -\frac{2}{y^3} dy$,so $\frac{dy}{y^3} = -\frac{dt}{2}$.
Let $u = \sqrt{x}$,then $du = \frac{dx}{2\sqrt{x}}$,so $\frac{dx}{\sqrt{x}} = 2du$.
Substituting these into the integral,we get $\int e^t (-\frac{1}{2}) dt = \int \cos u (2 du)$.
$-\frac{1}{2} e^t = 2 \sin u + C$.
Substituting back $t = \frac{1}{y^2}$ and $u = \sqrt{x}$,we have $-\frac{1}{2} e^{1/y^2} = 2 \sin \sqrt{x} + C$.
Given $y(0) = 1$,at $x = 0$,$y = 1$,so $-\frac{1}{2} e^1 = 2 \sin(0) + C$,which gives $C = -\frac{e}{2}$.
Thus,$-\frac{1}{2} e^{1/y^2} = 2 \sin \sqrt{x} - \frac{e}{2}$.
Multiplying by $-2$,we get $e^{1/y^2} = e - 4 \sin \sqrt{x}$.
Taking the natural logarithm on both sides,$\frac{1}{y^2} = \log_e(e - 4 \sin \sqrt{x})$.
Comparing this with $\frac{1}{y^2} = \log_e(f(x))$,we get $f(x) = e - 4 \sin \sqrt{x}$.

(B) Given differential equation is $2 x \frac{d y}{d x} - y = 4$.
Rearranging the terms,we get $2 x \frac{d y}{d x} = 4 + y$.
Separating the variables,we have $\frac{2}{4 + y} d y = \frac{1}{x} d x$.
Integrating both sides,$\int \frac{2}{4 + y} d y = \int \frac{1}{x} d x$.
This gives $2 \ln |4 + y| = \ln |x| + C$,where $C = \ln |c|$.
Using logarithmic properties,$\ln (4 + y)^2 = \ln |c x|$.
Taking the exponential of both sides,$(4 + y)^2 = c x$.
This equation is of the form $(y - k)^2 = 4 a (x - h)$,which represents a family of parabolas.

(A) Given differential equation: $e^x \tan y \, dx + (1+e^x) \sec^2 y \, dy = 0$
Rearranging the terms: $e^x \tan y \, dx = -(1+e^x) \sec^2 y \, dy$
Separating the variables: $\frac{e^x}{1+e^x} \, dx = -\frac{\sec^2 y}{\tan y} \, dy$
Integrating both sides: $\int \frac{e^x}{1+e^x} \, dx = -\int \frac{\sec^2 y}{\tan y} \, dy$
Using the formula $\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + C$,we get:
$\ln(1+e^x) = -\ln(\tan y) + \ln C$
$\ln(1+e^x) + \ln(\tan y) = \ln C$
$\ln[(1+e^x) \tan y] = \ln C$
$(1+e^x) \tan y = C$
Since the curve passes through the point $\left(0, \frac{\pi}{4}\right)$,substitute $x=0$ and $y=\frac{\pi}{4}$:
$(1+e^0) \tan(\frac{\pi}{4}) = C$
$(1+1)(1) = C \implies C = 2$
Thus,the equation of the curve is $(1+e^x) \tan y = 2$.

(C) Given the differential equation: $\frac{dy}{dx} = e^{x+y}$
Using the property of exponents,we can write: $\frac{dy}{dx} = e^x \cdot e^y$
Separating the variables,we get: $\frac{dy}{e^y} = e^x dx$
This can be rewritten as: $e^{-y} dy = e^x dx$
Integrating both sides: $\int e^{-y} dy = \int e^x dx$
Performing the integration: $-e^{-y} = e^x + C_1$
Rearranging the terms: $e^x + e^{-y} = -C_1$
Letting $-C_1 = c$,we get the final solution: $e^x + e^{-y} = c$

(D) Given the differential equation: $\frac{dy}{dx} = \sec y$.
Separating the variables,we get: $\cos y \, dy = dx$.
Integrating both sides: $\int \cos y \, dy = \int dx$.
This gives: $\sin y = x + c$.
Using the initial condition $y(0) = 0$,we substitute $x = 0$ and $y = 0$: $\sin(0) = 0 + c$,which implies $c = 0$.
Substituting $c = 0$ back into the general solution,we get: $\sin y = x$,or $x = \sin y$.
Thus,the correct option is $D$.

(C) Given the differential equation: $\frac{dy}{dx} = \frac{y(1+x)}{-x(1+y)}$.
Separate the variables $x$ and $y$:
$\frac{1+y}{y} dy = -\frac{1+x}{x} dx$.
This can be written as:
$(\frac{1}{y} + 1) dy = -(\frac{1}{x} + 1) dx$.
Integrate both sides:
$\int (\frac{1}{y} + 1) dy = -\int (\frac{1}{x} + 1) dx$.
$\log|y| + y = -(\log|x| + x) + C$.
$\log|y| + y = -\log|x| - x + C$.
Rearranging the terms:
$x + y + \log|x| + \log|y| = C$.
Using the property $\log a + \log b = \log(ab)$:
$x + y + \log|xy| = C$.
Thus,the correct option is $C$.

(A) Given differential equation is $(e^y + 1) \cos x \, dx + e^y \sin x \, dy = 0$.
Rearranging the terms,we get $\frac{e^y}{e^y + 1} \, dy = -\frac{\cos x}{\sin x} \, dx$.
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 $\int \frac{1}{u} \, du = -\ln|\sin x| + C$.
So,$\ln(e^y + 1) = -\ln|\sin x| + C$,which simplifies to $\ln(e^y + 1) + \ln|\sin x| = C$.
This gives $\ln((e^y + 1) \sin x) = C$,or $(e^y + 1) \sin x = K$ (where $K = e^C$).
The curve passes through $\left(\frac{\pi}{6}, 0\right)$,so substitute $x = \frac{\pi}{6}$ and $y = 0$:
$(e^0 + 1) \sin(\frac{\pi}{6}) = K \implies (1 + 1) \cdot \frac{1}{2} = K \implies K = 1$.
Thus,$(e^y + 1) \sin x = 1$,which means $e^y + 1 = \frac{1}{\sin x} = \operatorname{cosec} x$.
Therefore,$e^y = \operatorname{cosec} x - 1$,and $y = \log_e(\operatorname{cosec} x - 1)$.

(B) Given the differential equation $(x-y)^2 \frac{dy}{dx} = a^2$.
Let $v = x - y$. Then $\frac{dv}{dx} = 1 - \frac{dy}{dx}$,which implies $\frac{dy}{dx} = 1 - \frac{dv}{dx}$.
Substituting these into the equation: $v^2 (1 - \frac{dv}{dx}) = a^2$.
$1 - \frac{dv}{dx} = \frac{a^2}{v^2} \implies \frac{dv}{dx} = 1 - \frac{a^2}{v^2} = \frac{v^2 - a^2}{v^2}$.
Separating variables: $\frac{v^2}{v^2 - a^2} dv = dx$.
Integrating both sides: $\int \frac{v^2 - a^2 + a^2}{v^2 - a^2} dv = \int dx$.
$\int (1 + \frac{a^2}{v^2 - a^2}) dv = x + c$.
$v + a^2 \cdot \frac{1}{2a} \log \left| \frac{v-a}{v+a} \right| = x + c$.
$v + \frac{a}{2} \log \left| \frac{v-a}{v+a} \right| = x + c$.
Substituting $v = x - y$: $(x-y) + \frac{a}{2} \log \left| \frac{x-y-a}{x-y+a} \right| = x + c$.
$-y + \frac{a}{2} \log \left| \frac{x-y-a}{x-y+a} \right| = c$.
$y = \frac{a}{2} \log \left| \frac{x-y-a}{x-y+a} \right| + c'$.

(D) Given differential equation: $(2x - 4y + 3) \frac{dy}{dx} + (x - 2y + 1) = 0$.
Rearranging,we get $\frac{dy}{dx} = -\frac{x - 2y + 1}{2x - 4y + 3} = -\frac{(x - 2y) + 1}{2(x - 2y) + 3}$.
Let $v = x - 2y$. Then $\frac{dv}{dx} = 1 - 2 \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{1}{2} (1 - \frac{dv}{dx})$.
Substituting these into the equation: $\frac{1}{2} (1 - \frac{dv}{dx}) = -\frac{v + 1}{2v + 3}$.
$1 - \frac{dv}{dx} = -\frac{2v + 2}{2v + 3} \Rightarrow \frac{dv}{dx} = 1 + \frac{2v + 2}{2v + 3} = \frac{2v + 3 + 2v + 2}{2v + 3} = \frac{4v + 5}{2v + 3}$.
Separating variables: $\frac{2v + 3}{4v + 5} dv = dx$.
Multiply by $2$: $\frac{4v + 6}{4v + 5} dv = 2 dx$.
$\int (1 + \frac{1}{4v + 5}) dv = \int 2 dx$.
$v + \frac{1}{4} \log |4v + 5| = 2x + C$.
Substitute $v = x - 2y$: $(x - 2y) + \frac{1}{4} \log |4(x - 2y) + 5| = 2x + C$.
$\frac{1}{4} \log |4(x - 2y) + 5| = x + 2y + C$.
$\log |4(x - 2y) + 5| = 4x + 8y + C'$.

(B) Given,$\frac{dy}{dx} = \sin(x+y) \tan(x+y) - 1$.
Let $x+y = z$. Then,differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dz}{dx}$,so $\frac{dy}{dx} = \frac{dz}{dx} - 1$.
Substituting this into the differential equation:
$\frac{dz}{dx} - 1 = \sin z \tan z - 1$
$\frac{dz}{dx} = \sin z \tan z = \sin z \cdot \frac{\sin z}{\cos z} = \frac{\sin^2 z}{\cos z}$.
Rearranging the variables:
$\int \frac{\cos z}{\sin^2 z} dz = \int dx$.
Let $\sin z = t$,then $\cos z dz = dt$.
$\int \frac{1}{t^2} dt = x + c$
$- \frac{1}{t} = x + c$
$- \operatorname{cosec} z = x + c$
Substituting $z = x+y$ back:
$- \operatorname{cosec}(x+y) = x + c$
$x + \operatorname{cosec}(x+y) = c$ (where $c$ is an arbitrary constant).

(A) Given differential equation is $\frac{dy}{dx} + 1 = e^{x+y}$.
Let $x + y = z$.
Differentiating both sides with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dz}{dx}$.
Substituting this into the original equation,we have $\frac{dz}{dx} = e^z$.
Separating the variables,we get $e^{-z} dz = dx$.
Integrating both sides,$\int e^{-z} dz = \int dx$.
This gives $-e^{-z} = x + c$.
Substituting $z = x + y$ back,we get $-e^{-(x+y)} = x + c$.
Rearranging the terms,we get $x + e^{-(x+y)} + c = 0$.

(B) Given differential equation is $\frac{dy}{dx} = (\frac{x}{y})^{-1/3}$.
By simplifying the right side,we get $\frac{dy}{dx} = (\frac{y}{x})^{1/3} = \frac{y^{1/3}}{x^{1/3}}$.
Separating the variables,we have $y^{-1/3} dy = x^{-1/3} dx$.
Integrating both sides,we get $\int y^{-1/3} dy = \int x^{-1/3} dx$.
Applying 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 get $y^{2/3} = x^{2/3} + \frac{2}{3}C'$.
Let $c = \frac{2}{3}C'$,then $y^{2/3} - x^{2/3} = c$.

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

(D) Given the differential equation: $(x - (x+y) \log(x+y)) dx + x dy = 0$.
Rearranging the terms: $x dy = ((x+y) \log(x+y) - x) dx$.
$\frac{dy}{dx} = \frac{(x+y) \log(x+y) - x}{x}$.
Let $v = x+y$,then $dv = dx + dy$,so $dy = dv - dx$.
Substituting into the equation: $\frac{dv}{dx} - 1 = \frac{v \log v - x}{x} = \frac{v \log v}{x} - 1$.
$\frac{dv}{dx} = \frac{v \log v}{x}$.
Separating the variables: $\frac{dv}{v \log v} = \frac{dx}{x}$.
Integrating both sides: $\int \frac{dv}{v \log v} = \int \frac{dx}{x}$.
Let $u = \log v$,then $du = \frac{1}{v} dv$.
$\int \frac{du}{u} = \int \frac{dx}{x} \implies \log|u| = \log|x| + \log|c|$.
$u = cx \implies \log v = cx$.
Substituting $v = x+y$ back: $\log(x+y) = cx$.

(C) Given differential equation is $(2x-y)^2 dy - 2(2x-y)^2 dx - 2 dx = 0$.
Rearranging the terms: $(2x-y)^2 dy = [2(2x-y)^2 + 2] dx$.
So,$\frac{dy}{dx} = \frac{2(2x-y)^2 + 2}{(2x-y)^2} = 2 + \frac{2}{(2x-y)^2}$.
Let $v = 2x-y$. Then $\frac{dv}{dx} = 2 - \frac{dy}{dx}$,which implies $\frac{dy}{dx} = 2 - \frac{dv}{dx}$.
Substituting this into the equation: $2 - \frac{dv}{dx} = 2 + \frac{2}{v^2}$.
$-\frac{dv}{dx} = \frac{2}{v^2} \implies -v^2 dv = 2 dx$.
Integrating both sides: $-\int v^2 dv = \int 2 dx$.
$-\frac{v^3}{3} = 2x + c_1$.
$v^3 = -6x + c$ (where $c = -3c_1$).
Substituting $v = 2x-y$ back: $(2x-y)^3 = -6x + c$,which simplifies to $(2x-y)^3 + 6x = c$.

(C) Given differential equation: $\tan x \tan y \, dx + \cos^2 x \operatorname{cosec}^2 y \, dy = 0$
Divide the entire equation by $\cos^2 x \tan y$:
$\frac{\tan x}{\cos^2 x} \, dx + \frac{\operatorname{cosec}^2 y}{\tan y} \, dy = 0$
$\tan x \sec^2 x \, dx + \cot y \operatorname{cosec}^2 y \, dy = 0$
Integrating both sides:
$\int \tan x \sec^2 x \, dx + \int \cot y \operatorname{cosec}^2 y \, dy = C_1$
Let $u = \tan x$,then $du = \sec^2 x \, dx$. Let $v = \cot y$,then $dv = -\operatorname{cosec}^2 y \, dy$.
$\int u \, du - \int v \, dv = C_1$
$\frac{u^2}{2} - \frac{v^2}{2} = C_1$
$\tan^2 x - \cot^2 y = 2C_1 = C$
Thus,the general solution is $\tan^2 x - \cot^2 y = C$.

(C) Given the differential equation $\frac{dy}{dx} = \frac{2x - 4y - 5}{x - 2y + 2}$.
We can rewrite the numerator as $2(x - 2y + 2) - 9$.
So,$\frac{dy}{dx} = \frac{2(x - 2y + 2) - 9}{x - 2y + 2}$.
Let $t = x - 2y + 2$. Then $\frac{dt}{dx} = 1 - 2\frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{1}{2}(1 - \frac{dt}{dx})$.
Substituting this into the equation: $\frac{1}{2}(1 - \frac{dt}{dx}) = \frac{2t - 9}{t} = 2 - \frac{9}{t}$.
$1 - \frac{dt}{dx} = 4 - \frac{18}{t} \Rightarrow \frac{dt}{dx} = \frac{18}{t} - 3 = \frac{18 - 3t}{t}$.
Separating variables: $\frac{t}{18 - 3t} dt = dx \Rightarrow \frac{t}{3(6 - t)} dt = dx$.
Using partial fractions: $-\frac{1}{3} \int \frac{t - 6 + 6}{t - 6} dt = \int dx \Rightarrow -\frac{1}{3} \int (1 + \frac{6}{t - 6}) dt = x + C_1$.
$-\frac{1}{3} (t + 6 \ln|t - 6|) = x + C_1 \Rightarrow -t - 6 \ln|t - 6| = 3x + C$.
Substituting $t = x - 2y + 2$: $-(x - 2y + 2) - 6 \ln|x - 2y + 2 - 6| = 3x + C$.
$-x + 2y - 2 - 6 \ln|x - 2y - 4| = 3x + C \Rightarrow -4x + 2y - 6 \ln|x - 2y - 4| = C + 2$.
Dividing by $-2$: $2x - y + 3 \ln|x - 2y - 4| = k$.
Comparing with $2x - y + c \log(x - 2y - 4) = k$,we get $c = 3$.

(A) Given differential equation is $\cos(x+y) dy = dx$,which can be written as $\frac{dy}{dx} = \sec(x+y)$.
Let $x+y = t$. Then $1 + \frac{dy}{dx} = \frac{dt}{dx}$,so $\frac{dy}{dx} = \frac{dt}{dx} - 1$.
Substituting this into the equation: $\frac{dt}{dx} - 1 = \sec(t) \Rightarrow \frac{dt}{dx} = 1 + \sec(t) = \frac{1+\cos(t)}{\cos(t)}$.
Separating the variables: $\int \frac{\cos(t)}{1+\cos(t)} dt = \int dx$.
Using the identity $1+\cos(t) = 2\cos^2(t/2)$,we get $\int \frac{\cos(t)}{2\cos^2(t/2)} dt = \int dx$.
Since $\cos(t) = 2\cos^2(t/2) - 1$,the integral becomes $\int \frac{2\cos^2(t/2)-1}{2\cos^2(t/2)} dt = \int dx$.
This simplifies to $\int (1 - \frac{1}{2}\sec^2(t/2)) dt = \int dx$.
Integrating both sides: $t - \tan(t/2) = x + C$.
Substituting $t = x+y$: $(x+y) - \tan((x+y)/2) = x + C \Rightarrow y - \tan((x+y)/2) = C$.
Given $y(0) = 0$,we have $0 - \tan(0/2) = C \Rightarrow C = 0$.
Thus,the solution is $y = \tan \left(\frac{x+y}{2}\right)$.

(D) Given the differential equation $\frac{dy}{dx} - y \log_{e} 0.5 = 0$.
Rearranging the terms,we get $\frac{dy}{dx} = y \log_{e} 0.5$.
Separating the variables,we have $\frac{dy}{y} = (\log_{e} 0.5) dx$.
Integrating both sides,$\int \frac{dy}{y} = \int (\log_{e} 0.5) dx$,which gives $\ln y = (\log_{e} 0.5) x + C$.
Using the initial condition $y(0) = 1$,we substitute $x = 0$ and $y = 1$: $\ln 1 = (\log_{e} 0.5)(0) + C$,so $0 = 0 + C$,which implies $C = 0$.
Thus,$\ln y = (\log_{e} 0.5) x$.
Taking the exponential of both sides,$y = e^{(\log_{e} 0.5) x} = (e^{\log_{e} 0.5})^x = (0.5)^x$.
We are given that $y(x) \rightarrow k$ as $x \rightarrow \infty$.
Therefore,$k = \lim_{x \rightarrow \infty} (0.5)^x$.
Since $0.5 < 1$,as $x \rightarrow \infty$,$(0.5)^x \rightarrow 0$.
Hence,$k = 0$.