Variable separable type differential equations Questions in English

(A) Given the differential equation: $\frac{dy}{dx} + \frac{x}{y} = \frac{a}{y}$.
Multiply both sides by $y$: $y \frac{dy}{dx} + x = a$.
This is a variable separable form: $y dy = (a - x) dx$.
Integrate both sides: $\int y dy = \int (a - x) dx$.
$\frac{y^2}{2} = ax - \frac{x^2}{2} + C$,where $C$ is the constant of integration.
Multiply by $2$: $y^2 = 2ax - x^2 + 2C$.
Rearrange the terms: $x^2 - 2ax + y^2 = 2C$.
Complete the square for $x$: $(x^2 - 2ax + a^2) + y^2 = 2C + a^2$.
$(x - a)^2 + y^2 = a^2 + 2C$.
This represents a circle with center $(a, 0)$ and radius squared $r^2 = a^2 + 2C$.
Therefore,the radius $r = \sqrt{a^2 + 2C}$.

(C) Given the differential equation: $\frac{dy}{dx} = 2xye^{x^2}$.
Separate the variables $x$ and $y$:
$\frac{dy}{y} = 2x e^{x^2} dx$.
Integrate both sides:
$\int \frac{dy}{y} = \int 2x e^{x^2} dx$.
Let $u = x^2$,then $du = 2x dx$.
The integral becomes:
$\ln|y| = \int e^u du = e^u + C = e^{x^2} + C$.
Exponentiating both sides:
$|y| = e^{e^{x^2} + C} = e^C \cdot e^{e^{x^2}}$.
Let $c = \pm e^C$,then the general solution is:
$y = c e^{e^{x^2}}$.

(A) Given the differential equation $\frac{dy}{dx} = (x + 9y)^2$.
Let $v = x + 9y$.
Differentiating with respect to $x$,we get $\frac{dv}{dx} = 1 + 9\frac{dy}{dx}$.
Thus,$\frac{dy}{dx} = \frac{1}{9} \left(\frac{dv}{dx} - 1\right)$.
Substituting this into the original equation: $\frac{1}{9} \left(\frac{dv}{dx} - 1\right) = v^2$.
$\frac{dv}{dx} - 1 = 9v^2 \implies \frac{dv}{dx} = 1 + 9v^2$.
Separating the variables: $\frac{dv}{1 + 9v^2} = dx$.
Integrating both sides: $\int \frac{dv}{1 + (3v)^2} = \int dx$.
$\frac{1}{3} \tan^{-1}(3v) = x + C$.
$\tan^{-1}(3v) = 3x + 3C$.
$3v = \tan(3x + C_1)$,where $C_1 = 3C$.
Substituting $v = x + 9y$: $3(x + 9y) = \tan(3x + C_1) \implies 3x + 27y = \tan(3x + C_1)$.
Given $x = 0, y = \frac{1}{27}$: $3(0) + 27(\frac{1}{27}) = \tan(3(0) + C_1) \implies 1 = \tan(C_1)$.
So,$C_1 = \frac{\pi}{4}$.
The particular solution is $3x + 27y = \tan(3x + \frac{\pi}{4})$.
Note that $\tan(3x + \frac{\pi}{4}) = \tan[3(x + \frac{\pi}{12})]$.
Thus,the correct option is $A$.

(D) Given the differential equation: $3 e^x \tan y \, dx + (1 - e^x) \sec^2 y \, dy = 0$.
Rearranging the terms to separate variables: $\frac{\sec^2 y}{\tan y} \, dy = -\frac{3 e^x}{1 - e^x} \, dx$.
Integrating both sides: $\int \frac{\sec^2 y}{\tan y} \, dy = \int \frac{3 e^x}{e^x - 1} \, dx$.
Let $u = \tan y$,then $du = \sec^2 y \, dy$. The left side becomes $\ln|\tan y|$.
For the right side,let $v = e^x - 1$,then $dv = e^x \, dx$. The right side becomes $3 \ln|e^x - 1| + C$.
So,$\ln|\tan y| = 3 \ln|e^x - 1| + C = \ln|e^x - 1|^3 + C$.
This implies $\tan y = K(e^x - 1)^3$.
Given $y(1) = \frac{\pi}{4}$,we have $\tan(\frac{\pi}{4}) = K(e^1 - 1)^3$,so $1 = K(e - 1)^3$,which gives $K = \frac{1}{(e - 1)^3}$.
Substituting $K$ back: $\tan y = \frac{(e^x - 1)^3}{(e - 1)^3} = \left(\frac{e^x - 1}{e - 1}\right)^3$.
Since $(e^x - 1)^3 = -(1 - e^x)^3$ and $(e - 1)^3 = -(1 - e)^3$,we get $\tan y = \left(\frac{1 - e^x}{1 - e}\right)^3$.

(A) Given the differential equation: $\left(\frac{2+\sin x}{y+1}\right) \frac{dy}{dx} = -\cos x$.
Separating the variables,we get: $\frac{dy}{y+1} = -\frac{\cos x}{2+\sin x} dx$.
Integrating both sides: $\int \frac{dy}{y+1} = -\int \frac{\cos x}{2+\sin x} dx$.
Let $u = 2+\sin x$,then $du = \cos x dx$.
So,$\ln|y+1| = -\ln|2+\sin x| + C$.
This simplifies to $\ln|y+1| + \ln|2+\sin x| = C$,or $\ln|(y+1)(2+\sin x)| = C$.
Thus,$(y+1)(2+\sin x) = K$ (where $K = e^C$).
Using the initial condition $y(0)=1$: $(1+1)(2+\sin 0) = K \implies 2(2+0) = K \implies K = 4$.
So,$(y+1)(2+\sin x) = 4$.
Now,find $y\left(\frac{\pi}{2}\right)$:
$(y(\frac{\pi}{2})+1)(2+\sin(\frac{\pi}{2})) = 4$.
$(y(\frac{\pi}{2})+1)(2+1) = 4$.
$3(y(\frac{\pi}{2})+1) = 4$.
$y(\frac{\pi}{2})+1 = \frac{4}{3}$.
$y(\frac{\pi}{2}) = \frac{4}{3} - 1 = \frac{1}{3}$.

(B) Given the differential equation $\frac{dy}{dx} + \sin \left(\frac{x+y}{2}\right) = \sin \left(\frac{x-y}{2}\right)$.
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:
$\sin \left(\frac{x+y}{2}\right) - \sin \left(\frac{x-y}{2}\right) = 2 \cos \left(\frac{x}{2}\right) \sin \left(\frac{y}{2}\right)$.
Thus,the equation becomes $\frac{dy}{dx} = -2 \cos \left(\frac{x}{2}\right) \sin \left(\frac{y}{2}\right)$.
Separating the variables,we get $\frac{dy}{\sin(y/2)} = -2 \cos(x/2) dx$.
Integrating both sides: $\int \csc(y/2) dy = -2 \int \cos(x/2) dx$.
$2 \log |\tan(y/4)| = -2(2 \sin(x/2)) + c_1$.
Dividing by $2$: $\log |\tan(y/4)| = -2 \sin(x/2) + c$,where $c = c_1/2$.
Therefore,the general solution is $\log \tan \left(\frac{y}{4}\right) = c - 2 \sin \left(\frac{x}{2}\right)$.

(B) Given the slope of the tangent 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,$\int \frac{dy}{y-1} = \int (\frac{1}{x} - \frac{1}{x+1}) dx$.
$\ln|y-1| = \ln|x| - \ln|x+1| + C$.
$\ln|y-1| = \ln|\frac{x}{x+1}| + C$.
Since the curve passes through $(1,0)$,substitute $x=1$ and $y=0$:
$\ln|0-1| = \ln|\frac{1}{1+1}| + C \implies 0 = \ln(\frac{1}{2}) + C \implies C = \ln(2)$.
Thus,$\ln|y-1| = \ln|\frac{x}{x+1}| + \ln(2) = \ln|\frac{2x}{x+1}|$.
$y-1 = \frac{2x}{x+1} \implies (y-1)(x+1) = 2x$.
Rearranging gives $2x - (y-1)(x+1) = 0$.

(C) Given the differential equation: $\frac{dy}{dx} = \cot x \cdot \cot y$.
Separating the variables,we get: $\frac{dy}{\cot y} = \cot x \cdot dx$.
This can be written as: $\tan y \cdot dy = \cot x \cdot dx$.
Integrating both sides: $\int \tan y \cdot dy = \int \cot x \cdot dx$.
Using the standard integration formulas $\int \tan y \cdot dy = \ln|\sec y|$ and $\int \cot x \cdot dx = \ln|\sin x|$,we get: $\ln|\sec y| = \ln|\sin x| + \ln|c|$.
Using the property $\ln|a| + \ln|b| = \ln|ab|$,we have: $\ln|\sec y| = \ln|c \sin x|$.
Taking the exponential of both sides: $\sec y = c \sin x$.
Since $\sec y = \frac{1}{\cos y}$,we have $\frac{1}{\cos y} = c \sin x$,which implies $\sin x = \frac{1}{c} \cos y$.
Letting $k = \frac{1}{c}$,we get $\sin x = k \cos y$.
Comparing this with the given options,option $C$ is the correct form.

(B) Given the differential equation $\log \left(\frac{dy}{dx}\right) = 2x - 5y$.
By definition of logarithm,we have $\frac{dy}{dx} = e^{2x - 5y} = e^{2x} \cdot e^{-5y}$.
Separating the variables,we get $e^{5y} \, dy = e^{2x} \, dx$.
Integrating both sides,$\int e^{5y} \, dy = \int e^{2x} \, dx$.
This gives $\frac{e^{5y}}{5} = \frac{e^{2x}}{2} + C$.
Multiplying by $10$,we get $2e^{5y} = 5e^{2x} + 10C$,or $2e^{5y} - 5e^{2x} = K$.
Using the initial condition $y(0) = 0$,we substitute $x = 0$ and $y = 0$:
$2e^{5(0)} - 5e^{2(0)} = K \implies 2(1) - 5(1) = K \implies K = -3$.
Thus,$2e^{5y} - 5e^{2x} = -3$,which can be rewritten as $5e^{2x} - 2e^{5y} = 3$.

(C) Given the differential equation $\frac{dy}{dx} = \frac{1}{x+y+1}$.
Let $v = x+y+1$. Then,differentiating with respect to $x$,we get $\frac{dv}{dx} = 1 + \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$.
Substituting these into the original equation: $\frac{dv}{dx} - 1 = \frac{1}{v}$.
Rearranging the terms: $\frac{dv}{dx} = 1 + \frac{1}{v} = \frac{v+1}{v}$.
Separating the variables: $\frac{v}{v+1} dv = dx$.
Integrating both sides: $\int \frac{v+1-1}{v+1} dv = \int dx$.
This simplifies to $\int (1 - \frac{1}{v+1}) dv = \int dx$.
Integrating gives $v - \log|v+1| = x + c$.
Substituting $v = x+y+1$ back into the equation: $(x+y+1) - \log|x+y+1+1| = x + c$.
This simplifies to $y+1 - \log|x+y+2| = c$,or $y = \log|x+y+2| + C'$,where $C' = c-1$ is a constant.
Thus,the correct option is $C$.

(D) Given differential equation is $\frac{dy}{dx}=(x-y)^2$ $(i)$.
Let $x-y=t$. Then $1-\frac{dy}{dx}=\frac{dt}{dx}$,which implies $\frac{dy}{dx}=1-\frac{dt}{dx}$.
Substituting into $(i)$,we get $1-\frac{dt}{dx}=t^2$,so $\frac{dt}{dx}=1-t^2$.
Separating variables,$dx = \frac{1}{1-t^2} dt$.
Integrating both sides,$x = \int \frac{1}{1-t^2} dt = \frac{1}{2} \log \left|\frac{1+t}{1-t}\right| + c$.
Substituting $t=x-y$,we get $x = \frac{1}{2} \log \left|\frac{1+x-y}{1-x+y}\right| + c$.
Given $y(1)=1$,at $x=1$,$y=1$: $1 = \frac{1}{2} \log \left|\frac{1+1-1}{1-1+1}\right| + c \Rightarrow 1 = \frac{1}{2} \log(1) + c \Rightarrow c=1$.
Thus,$x = \frac{1}{2} \log \left|\frac{1+x-y}{1-x+y}\right| + 1$.
$x-1 = \frac{1}{2} \log \left|\frac{1+x-y}{1-x+y}\right| \Rightarrow 2(x-1) = \log \left|\frac{1+x-y}{1-x+y}\right|$.
Using $\log(a/b) = -\log(b/a)$,we get $2(x-1) = -\log \left|\frac{1-x+y}{1+x-y}\right|$.

(D) Given the differential equation: $(2+\sin x) \frac{dy}{dx} + (y+1) \cos x = 0$.
Rearranging the terms to separate the variables: $\frac{1}{y+1} dy = -\frac{\cos x}{2+\sin x} dx$.
Integrating both sides: $\int \frac{1}{y+1} dy = -\int \frac{\cos x}{2+\sin x} dx$.
This gives: $\ln(y+1) = -\ln(2+\sin x) + C$.
Using the initial condition $y(0)=1$: $\ln(1+1) = -\ln(2+\sin 0) + C \implies \ln 2 = -\ln 2 + C \implies C = 2\ln 2 = \ln 4$.
Substituting $C$ back into the equation: $\ln(y+1) = -\ln(2+\sin x) + \ln 4 = \ln\left(\frac{4}{2+\sin x}\right)$.
Taking the exponential of both sides: $y+1 = \frac{4}{2+\sin x} \implies y = \frac{4}{2+\sin x} - 1$.
Now,evaluating at $x = \frac{\pi}{2}$: $y\left(\frac{\pi}{2}\right) = \frac{4}{2+\sin(\pi/2)} - 1 = \frac{4}{2+1} - 1 = \frac{4}{3} - 1 = \frac{1}{3}$.

(A) Given the differential equation: $\left(\frac{5+e^x}{2+y}\right) \frac{dy}{dx} + e^x = 0$.
Separating the variables,we get: $\frac{dy}{2+y} = -\frac{e^x}{5+e^x} dx$.
Integrating both sides: $\int \frac{dy}{2+y} = -\int \frac{e^x}{5+e^x} dx$.
This gives: $\log |2+y| = -\log |5+e^x| + C$.
Using the initial condition $y(0)=1$: $\log |2+1| = -\log |5+e^0| + C \Rightarrow \log 3 = -\log 6 + C \Rightarrow C = \log 3 + \log 6 = \log 18$.
Thus,$\log |2+y| = \log \left|\frac{18}{5+e^x}\right|$,which implies $2+y = \frac{18}{5+e^x}$.
So,$y(x) = \frac{18}{5+e^x} - 2$.
For $x = \log 13$,$y(\log 13) = \frac{18}{5+e^{\log 13}} - 2 = \frac{18}{5+13} - 2 = \frac{18}{18} - 2 = 1 - 2 = -1$.

(C) The equation of the given circle is $x^2+y^2-2x-2y=0$.
Completing the square,we get $(x-1)^2+(y-1)^2=2$.
Thus,the centre of the circle is $(1, 1)$.
The slope of the tangent to the curve is given by $\frac{dy}{dx} = \frac{2y}{x^2}$.
Separating the variables,we have $\int \frac{1}{y} dy = \int \frac{2}{x^2} dx$.
Integrating both sides,we get $\log |y| = -\frac{2}{x} + c$.
Since the curve passes through $(1, 1)$,we substitute $x=1$ and $y=1$:
$\log |1| = -\frac{2}{1} + c \implies 0 = -2 + c \implies c = 2$.
Substituting $c=2$ into the general solution,we get $\log |y| = -\frac{2}{x} + 2$.
Multiplying by $x$,we get $x \log |y| = -2 + 2x = 2(x-1)$.

(B) Given differential equation: $(1+y^2)(1+\log x) dx + x dy = 0$
Rearranging the terms: $(1+y^2)(1+\log x) dx = -x dy$
Separating the variables: $\frac{1+\log x}{x} dx = -\frac{1}{1+y^2} dy$
Integrating both sides: $\int \frac{1+\log x}{x} dx = -\int \frac{1}{1+y^2} dy$
Let $1+\log x = t$,then $\frac{1}{x} dx = dt$.
Substituting this: $\int t dt = -\tan^{-1} y + C$
$\frac{t^2}{2} = -\tan^{-1} y + C$
$\frac{(1+\log x)^2}{2} = -\tan^{-1} y + C$
At $x=1, y=1$: $\frac{(1+\log 1)^2}{2} = -\tan^{-1}(1) + C$
$\frac{1}{2} = -\frac{\pi}{4} + C \implies C = \frac{1}{2} + \frac{\pi}{4}$
Substituting $C$ back: $\frac{(1+\log x)^2}{2} = -\tan^{-1} y + \frac{1}{2} + \frac{\pi}{4}$
$\frac{1 + 2\log x + (\log x)^2}{2} = -\tan^{-1} y + \frac{1}{2} + \frac{\pi}{4}$
$\frac{1}{2} + \log x + \frac{(\log x)^2}{2} = -\tan^{-1} y + \frac{1}{2} + \frac{\pi}{4}$
$\log x + \frac{(\log x)^2}{2} + \tan^{-1} y = \frac{\pi}{4}$

(D) Given the differential equation: $\frac{1}{x} \frac{dy}{dx} = \tan^{-1} x$
Separating the variables,we get: $dy = x \tan^{-1} x dx$
Integrating both sides: $y = \int x \tan^{-1} x dx$
Using integration by parts,$\int u dv = uv - \int v du$,where $u = \tan^{-1} x$ and $dv = x dx$:
$y = \tan^{-1} x \cdot \frac{x^2}{2} - \int \frac{1}{1+x^2} \cdot \frac{x^2}{2} dx$
$y = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} \int \frac{x^2}{1+x^2} dx$
Adding and subtracting $1$ in the numerator:
$y = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} \int \frac{x^2+1-1}{1+x^2} dx$
$y = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} \left( \int 1 dx - \int \frac{1}{1+x^2} dx \right)$
$y = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} (x - \tan^{-1} x) + c$

(A) Given the differential equation: $\frac{dy}{dx} = \frac{x+y+1}{x+y-1} \dots (i)$
Let $x+y = v$. Then,differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dv}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1 \dots (ii)$
Substituting $(ii)$ into $(i)$:
$\frac{dv}{dx} - 1 = \frac{v+1}{v-1}$
$\frac{dv}{dx} = \frac{v+1}{v-1} + 1 = \frac{v+1+v-1}{v-1} = \frac{2v}{v-1}$
Separating the variables:
$\frac{v-1}{2v} dv = dx$
$\frac{1}{2} (1 - \frac{1}{v}) dv = dx$
Integrating both sides:
$\frac{1}{2} (v - \log|v|) = x + c_1$
$v - \log|v| = 2x + 2c_1$
Substituting $v = x+y$:
$(x+y) - \log|x+y| = 2x + c$
$y - \log|x+y| = x + c$
$y = x + \log|x+y| + c$,where $c = -2c_1$.

(A) Given differential equation: $e^{y-x} \frac{dy}{dx} = y \left( \frac{\sin x + \cos x}{1 + y \log y} \right)$
Rearranging the terms: $\frac{e^y}{e^x} \frac{dy}{dx} = \frac{y}{1 + y \log y} (\sin x + \cos x)$
Separating the variables: $\frac{e^y (1 + y \log y)}{y} dy = e^x (\sin x + \cos x) dx$
Simplifying the left side: $e^y \left( \log y + \frac{1}{y} \right) dy = e^x (\sin x + \cos x) dx$
Integrating both sides: $\int e^y \left( \log y + \frac{1}{y} \right) dy = \int e^x (\sin x + \cos x) dx$
Using the identity $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$,we get: $e^y \log y = e^x \sin x + c$.

(B) Given the differential equation: $\frac{dy}{dx} + \sin \left(\frac{x+y}{2}\right) = \sin \left(\frac{x-y}{2}\right)$
Rearranging the terms: $\frac{dy}{dx} = \sin \left(\frac{x-y}{2}\right) - \sin \left(\frac{x+y}{2}\right)$
Using the trigonometric identity $\sin A - \sin B = 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$,we get:
$\frac{dy}{dx} = 2 \cos \left(\frac{x}{2}\right) \sin \left(-\frac{y}{2}\right) = -2 \sin \left(\frac{y}{2}\right) \cos \left(\frac{x}{2}\right)$
Separating the variables: $\int \operatorname{cosec} \left(\frac{y}{2}\right) dy = -\int 2 \cos \left(\frac{x}{2}\right) dx$
Integrating both sides: $2 \log \tan \left(\frac{y}{4}\right) = -4 \sin \left(\frac{x}{2}\right) + c_1$
Dividing by $2$: $\log \tan \left(\frac{y}{4}\right) = -2 \sin \left(\frac{x}{2}\right) + C$,where $C = \frac{c_1}{2}$.

(A) Given the differential equation: $(x+2) \frac{dy}{dx} = x^2+4x-9$.
We can rewrite the right side as: $x^2+4x-9 = (x^2+4x+4) - 13 = (x+2)^2 - 13$.
So,$\frac{dy}{dx} = \frac{(x+2)^2 - 13}{x+2} = (x+2) - \frac{13}{x+2}$.
Integrating both sides with respect to $x$:
$\int dy = \int (x+2) dx - 13 \int \frac{1}{x+2} dx$.
$y = \frac{(x+2)^2}{2} - 13 \ln|x+2| + C$.
Given $y(0) = 0$,substitute $x=0$ and $y=0$:
$0 = \frac{(0+2)^2}{2} - 13 \ln|0+2| + C$.
$0 = 2 - 13 \ln(2) + C \implies C = 13 \ln(2) - 2$.
Thus,the solution is $y(x) = \frac{(x+2)^2}{2} - 13 \ln|x+2| + 13 \ln(2) - 2$.
Now,find $y(-4)$:
$y(-4) = \frac{(-4+2)^2}{2} - 13 \ln|-4+2| + 13 \ln(2) - 2$.
$y(-4) = \frac{(-2)^2}{2} - 13 \ln(2) + 13 \ln(2) - 2$.
$y(-4) = \frac{4}{2} - 2 = 2 - 2 = 0$.

(C) Given the differential equation: $\frac{dy}{dx} = \frac{3e^{2x} + 3e^{4x}}{e^x + e^{-x}}$
Simplify the numerator by factoring out $3e^{2x}$: $3e^{2x}(1 + e^{2x})$
Simplify the denominator: $e^x + e^{-x} = e^x + \frac{1}{e^x} = \frac{e^{2x} + 1}{e^x}$
Substitute these back into the equation: $\frac{dy}{dx} = \frac{3e^{2x}(1 + e^{2x})}{\frac{e^{2x} + 1}{e^x}}$
Cancel the common term $(1 + e^{2x})$: $\frac{dy}{dx} = 3e^{2x} \cdot e^x = 3e^{3x}$
Integrate both sides with respect to $x$: $\int dy = \int 3e^{3x} dx$
$y = 3 \cdot \frac{e^{3x}}{3} + c = e^{3x} + c$

(C) Given differential equation is $(1+y^2) dx - xy dy = 0$.
Rearranging the terms,we get $(1+y^2) dx = xy dy$.
Separating the variables,we have $\frac{1}{x} dx = \frac{y}{1+y^2} dy$.
Integrating both sides: $\int \frac{1}{x} dx = \int \frac{y}{1+y^2} dy$.
This gives $\log |x| = \frac{1}{2} \log (1+y^2) + C$.
Given $x=1$ and $y=0$,we substitute these values: $\log(1) = \frac{1}{2} \log(1+0^2) + C$,which implies $0 = 0 + C$,so $C=0$.
The equation becomes $\log x = \frac{1}{2} \log (1+y^2)$.
Multiplying by $2$,we get $2 \log x = \log (1+y^2)$,which is $\log(x^2) = \log(1+y^2)$.
Taking the exponential of both sides,$x^2 = 1+y^2$,or $x^2 - y^2 = 1$.
This equation represents a rectangular hyperbola.

(C) Given differential equation is $\log \left(\frac{d y}{d x}\right)=a x+b y$.
Taking exponential on both sides,we get $\frac{d y}{d x}=e^{a x+b y} = e^{a x} \cdot e^{b y}$.
Separating the variables,we have $\frac{d y}{e^{b y}} = e^{a x} d x$,which is $e^{-b y} d y = e^{a x} d x$.
Integrating both sides,we get $\int e^{-b y} d y = \int e^{a x} d x$.
This results in $\frac{e^{-b y}}{-b} = \frac{e^{a x}}{a} + C$.
Rearranging the terms,we get $\frac{e^{a x}}{a} + \frac{e^{-b y}}{b} = -C$.
Multiplying by $ab$,we get $b e^{a x} + a e^{-b y} = -abC$.
Letting $c_1 = -abC$,we obtain $a e^{-b y} + b e^{a x} = c_1$.

(C) Given differential equation is $\cos x(1+\cos y) dx - \sin y(1+\sin x) dy = 0$.
Rearranging the terms to separate the variables,we get:
$\cos x(1+\cos y) dx = \sin y(1+\sin x) dy$
$\frac{\cos x}{1+\sin x} dx = \frac{\sin y}{1+\cos y} dy$
Integrating both sides:
$\int \frac{\cos x}{1+\sin x} dx = \int \frac{\sin y}{1+\cos y} dy$
Let $u = 1+\sin x$,then $du = \cos x dx$.
Let $v = 1+\cos y$,then $dv = -\sin y dy$,so $\sin y dy = -dv$.
Substituting these into the integral:
$\int \frac{1}{u} du = \int -\frac{1}{v} dv$
$\ln|u| = -\ln|v| + \ln|c|$
$\ln|1+\sin x| = -\ln|1+\cos y| + \ln|c|$
$\ln|1+\sin x| + \ln|1+\cos y| = \ln|c|$
$\ln|(1+\sin x)(1+\cos y)| = \ln|c|$
Taking the exponential of both sides:
$(1+\sin x)(1+\cos y) = c$.

(B) Given the differential equation: $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$.
By separating the variables,we get: $\frac{dy}{1+y^2} = \frac{dx}{1+x^2}$.
Integrating both sides: $\int \frac{dy}{1+y^2} = \int \frac{dx}{1+x^2}$.
This yields: $\tan^{-1}(y) = \tan^{-1}(x) + C$,where $C$ is a constant of integration.
Rearranging the terms: $\tan^{-1}(y) - \tan^{-1}(x) = C$.
Using the formula $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}(\frac{A-B}{1+AB})$,we get: $\tan^{-1}(\frac{y-x}{1+xy}) = C$.
Taking the tangent of both sides: $\frac{y-x}{1+xy} = \tan(C)$.
Let $\tan(C) = c$,where $c$ is a new constant.
Thus,$y-x = c(1+xy)$.

(B) Given the differential equation $\frac{dy}{dx} = y + 3$.
Separating the variables,we get $\frac{dy}{y + 3} = dx$.
Integrating both sides,we have $\int \frac{dy}{y + 3} = \int dx + C$.
This gives $\log|y + 3| = x + C$.
Given the initial condition $y(0) = 2$,substituting $x = 0$ and $y = 2$ into the equation:
$\log|2 + 3| = 0 + C \implies C = \log 5$.
Thus,the equation becomes $\log(y + 3) = x + \log 5$.
Taking the exponential of both sides,$y + 3 = e^{x + \log 5} = 5e^x$.
So,$y = 5e^x - 3$.
Now,calculating $y(\log 2)$:
$y(\log 2) = 5e^{\log 2} - 3$.
Since $e^{\log 2} = 2$,we get $y(\log 2) = 5(2) - 3 = 10 - 3 = 7$.

(C) Given the differential equation: $\cos (x+y) dy = dx$.
Dividing by $dy$,we get $\frac{dx}{dy} = \cos (x+y)$.
Let $x+y = u$. Differentiating with respect to $y$,we get $\frac{dx}{dy} + 1 = \frac{du}{dy}$,which implies $\frac{dx}{dy} = \frac{du}{dy} - 1$.
Substituting this into the differential equation: $\frac{du}{dy} - 1 = \cos u$.
Rearranging the terms: $\frac{du}{dy} = 1 + \cos u$.
Separating the variables: $\frac{du}{1 + \cos u} = dy$.
Using the identity $1 + \cos u = 2 \cos^2 \left(\frac{u}{2}\right)$,we get $\frac{du}{2 \cos^2 \left(\frac{u}{2}\right)} = dy$,which simplifies to $\frac{1}{2} \sec^2 \left(\frac{u}{2}\right) du = dy$.
Integrating both sides: $\int \frac{1}{2} \sec^2 \left(\frac{u}{2}\right) du = \int dy$.
This gives $\tan \left(\frac{u}{2}\right) = y + c$.
Substituting $u = x+y$ back,we get $\tan \left(\frac{x+y}{2}\right) = y + c$,or $y = \tan \left(\frac{x+y}{2}\right) - c$. Since $c$ is an arbitrary constant,we can write the general solution as $y = \tan \left(\frac{x+y}{2}\right) + c$.

(B) Given the differential equation: $e^{\frac{dy}{dx}} = (x+1)$.
Taking the natural logarithm on both sides,we get: $\frac{dy}{dx} = \log(x+1)$.
Integrating both sides with respect to $x$: $\int dy = \int \log(x+1) dx + C$.
Using integration by parts for $\int \log(x+1) dx$ (let $u = \log(x+1)$ and $dv = dx$):
$y = (x+1) \log(x+1) - \int (x+1) \cdot \frac{1}{x+1} dx + C$.
$y = (x+1) \log(x+1) - \int 1 dx + C$.
$y = (x+1) \log(x+1) - x + C$.
Given the 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 have $3 = 0 + C$,so $C = 3$.
Therefore,the particular solution is $y = (x+1) \log(x+1) - x + 3$.

(C) Given differential equation is $\cos x(1+\cos y) dx - \sin y(1+\sin x) dy = 0$.
Rearranging the terms to separate the variables,we get:
$\frac{\cos x}{1+\sin x} dx = \frac{\sin y}{1+\cos y} dy$.
Integrating both sides:
$\int \frac{\cos x}{1+\sin x} dx = \int \frac{\sin y}{1+\cos y} dy$.
Let $u = 1+\sin x$,then $du = \cos x dx$.
Let $v = 1+\cos y$,then $dv = -\sin y dy$.
Substituting these into the integral:
$\int \frac{1}{u} du = -\int \frac{1}{v} dv$.
$\ln |u| = -\ln |v| + \ln |c|$.
$\ln |u| + \ln |v| = \ln |c|$.
$\ln |uv| = \ln |c|$.
$uv = c$.
Substituting back the values of $u$ and $v$:
$(1+\sin x)(1+\cos y) = c$.

(B) Given the differential equation $\sin^3 x \frac{dx}{dy} = \sin y$.
Separating the variables,we get $\int \sin^3 x \, dx = \int \sin y \, dy$.
Using the trigonometric identity $\sin 3x = 3 \sin x - 4 \sin^3 x$,we have $\sin^3 x = \frac{3 \sin x - \sin 3x}{4}$.
Substituting this into the integral,we get $\int \frac{3 \sin x - \sin 3x}{4} \, dx = \int \sin y \, dy$.
Integrating both sides,we get $\frac{3}{4} (-\cos x) - \frac{1}{4} (-\frac{\cos 3x}{3}) = -\cos y + C$.
This simplifies to $-\frac{3}{4} \cos x + \frac{1}{12} \cos 3x = -\cos y + C$.
Rearranging the terms,we get $\cos y - \frac{3}{4} \cos x + \frac{1}{12} \cos 3x = C$.

(B) Given the differential equation: $\frac{2+\sin x}{y+1} \frac{d y}{d x} = -\cos x$
Separate the variables: $\int \frac{d y}{y+1} = \int \frac{-\cos x}{2+\sin x} d x$
Integrate both sides: $\ln|y+1| = -\ln|2+\sin x| + C_1$
This simplifies to: $\ln|y+1| + \ln|2+\sin x| = C_1$
Using logarithmic properties: $\ln|(y+1)(2+\sin x)| = C_1$
Exponentiating both sides: $(y+1)(2+\sin x) = C$
Given $y(0)=1$,substitute $x=0$ and $y=1$: $(1+1)(2+\sin 0) = C \Rightarrow 2(2+0) = C \Rightarrow C=4$
So,the equation is: $(y+1)(2+\sin x) = 4$
To find $y\left(\frac{\pi}{2}\right)$,substitute $x=\frac{\pi}{2}$: $(y+1)(2+\sin\frac{\pi}{2}) = 4$
$(y+1)(2+1) = 4 \Rightarrow 3(y+1) = 4 \Rightarrow y+1 = \frac{4}{3}$
$y = \frac{4}{3} - 1 = \frac{1}{3}$

(A) Given the differential equation: $\frac{dy}{dx} = e^{x+y} + x^2 e^{x^3+y}$
Separating the terms: $\frac{dy}{dx} = e^y(e^x + x^2 e^{x^3})$
Rearranging the variables: $e^{-y} dy = (e^x + x^2 e^{x^3}) dx$
Integrating both sides: $\int e^{-y} dy = \int (e^x + x^2 e^{x^3}) dx$
Solving the integrals: $-e^{-y} = e^x + \frac{1}{3} e^{x^3} + C$
Rearranging to standard form: $e^{-y} + e^x + \frac{1}{3} e^{x^3} = -C$
Since $-C$ is also a constant,we can write: $e^{-y} + e^x + \frac{1}{3} e^{x^3} = C$

(D) Given the differential equation: $\frac{dy}{dx} = e^{2y} \cos x$.
Separating the variables,we get: $e^{-2y} dy = \cos x dx$.
Integrating both sides: $\int e^{-2y} dy = \int \cos x dx$.
This gives: $-\frac{1}{2} e^{-2y} = \sin x + C$.
Given the condition $y(\frac{\pi}{6}) = 0$,substitute $x = \frac{\pi}{6}$ and $y = 0$:
$-\frac{1}{2} e^{0} = \sin(\frac{\pi}{6}) + C$.
$-\frac{1}{2} = \frac{1}{2} + C$,which implies $C = -1$.
Substituting $C$ back into the equation: $-\frac{1}{2} e^{-2y} = \sin x - 1$.
Multiplying by $-2$: $e^{-2y} = -2 \sin x + 2$.
Rearranging the terms: $2 \sin x + e^{-2y} - 2 = 0$.

(C) Given differential equation is $(y^3+y)(x^2+1) dy = x(y^4+2y^2) dx$.
Separating the variables,we get:
$\frac{y^3+y}{y^4+2y^2} dy = \frac{x}{x^2+1} dx$.
Multiply the numerator and denominator of the left side by $2$:
$\frac{1}{2} \int \frac{2y^3+2y}{y^4+2y^2} dy = \int \frac{x}{x^2+1} dx$.
Let $u = y^4+2y^2$,then $du = (4y^3+4y) dy = 2(2y^3+2y) dy$,so $(2y^3+2y) dy = \frac{1}{2} du$.
Substituting this:
$\frac{1}{2} \int \frac{1}{2u} du = \int \frac{x}{x^2+1} dx$.
$\frac{1}{4} \ln|y^4+2y^2| = \frac{1}{2} \ln|x^2+1| + \ln|C_1|$.
Multiply by $4$:
$\ln|y^4+2y^2| = 2 \ln|x^2+1| + 4 \ln|C_1| = \ln|(x^2+1)^2| + \ln|C_1^4|$.
Taking exponential on both sides:
$y^4+2y^2 = C(x^2+1)^2$,where $C = C_1^4$.
$y^2(y^2+2) = C(x^2+1)^2$.

(B) Given the differential equation $\log \left(\frac{dy}{dx}\right) = 3x + 4y$.
Taking the exponential on both sides,we get $\frac{dy}{dx} = e^{3x + 4y} = e^{3x} \cdot e^{4y}$.
Separating the variables,we have $e^{-4y} \, dy = e^{3x} \, dx$.
Integrating both sides,$\int e^{-4y} \, dy = \int e^{3x} \, dx$.
This gives $\frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} + C$.
Given the initial condition $x = 0$ and $y = 0$,we substitute these values into the equation:
$\frac{e^{0}}{-4} = \frac{e^{0}}{3} + C \Rightarrow -\frac{1}{4} = \frac{1}{3} + C$.
$C = -\frac{1}{4} - \frac{1}{3} = -\frac{7}{12}$.
Substituting $C$ back into the equation: $\frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} - \frac{7}{12}$.
Multiplying the entire equation by $-12$: $3e^{-4y} = -4e^{3x} + 7$.
Rearranging the terms,we get $3e^{-4y} + 4e^{3x} = 7$.

(D) Given the differential equation: $(1+e^{-x})(1+y^2) \frac{dy}{dx} = y^2$
Separate the variables: $\frac{1+y^2}{y^2} dy = \frac{1}{1+e^{-x}} dx$
Since $\frac{1}{1+e^{-x}} = \frac{e^x}{e^x+1}$,we have: $\int (y^{-2} + 1) dy = \int \frac{e^x}{e^x+1} dx$
Integrating both sides: $-\frac{1}{y} + y = \log(1+e^x) + C$
Multiply by $y$: $y^2 - 1 = y \log(1+e^x) + Cy$
$y^2 - 1 = y(\log(1+e^x) + C)$
Given the curve passes through $(0,1)$,substitute $x=0, y=1$:
$1^2 - 1 = 1(\log(1+e^0) + C) \Rightarrow 0 = \log(2) + C \Rightarrow C = -\log(2)$
Substitute $C$ back into the equation: $y^2 - 1 = y(\log(1+e^x) - \log(2))$
$y^2 - 1 = y \log(\frac{1+e^x}{2})$
$y^2 = 1 + y \log(\frac{1+e^x}{2})$

(D) Given the differential equation: $x^2(y+1) dx + y^2(x-1) dy = 0$.
Rearranging the terms to separate variables: $\frac{x^2}{x-1} dx + \frac{y^2}{y+1} dy = 0$.
Integrating both sides: $\int \frac{x^2}{x-1} dx + \int \frac{y^2}{y+1} dy = C'$.
Using polynomial division: $\frac{x^2}{x-1} = x+1 + \frac{1}{x-1}$ and $\frac{y^2}{y+1} = y-1 + \frac{1}{y+1}$.
Substituting these into the integral: $\int (x+1 + \frac{1}{x-1}) dx + \int (y-1 + \frac{1}{y+1}) dy = C'$.
Integrating term by term: $(\frac{x^2}{2} + x + \log |x-1|) + (\frac{y^2}{2} - y + \log |y+1|) = C'$.
Multiplying by $2$: $x^2 + 2x + y^2 - 2y + 2 \log |(x-1)(y+1)| = 2C'$.
Completing the squares: $(x^2 + 2x + 1) + (y^2 - 2y + 1) + 2 \log |(x-1)(y+1)| = 2C' + 2$.
$(x+1)^2 + (y-1)^2 + 2 \log |(x-1)(y+1)| = C$.

(D) Given the differential equation: $\frac{dy}{dx} = 1 + x + y^2 + xy^2$
Factorizing the right side: $\frac{dy}{dx} = (1 + x)(1 + y^2)$
Separating the variables: $\int \frac{dy}{1 + y^2} = \int (1 + x) dx$
Integrating both sides: $\tan^{-1}(y) = x + \frac{x^2}{2} + C$
Given the initial condition $y(0) = 0$: $\tan^{-1}(0) = 0 + \frac{0^2}{2} + C \Rightarrow C = 0$
Thus,the particular solution is: $\tan^{-1}(y) = x + \frac{x^2}{2}$
Therefore: $y = \tan \left(x + \frac{x^2}{2}\right)$

(D) Given the differential equation: $\frac{dy}{dx} = 1 - x + y - xy$
Factor the right side: $\frac{dy}{dx} = (1 - x) + y(1 - x) = (1 - x)(1 + y)$
Separate the variables: $\frac{dy}{1 + y} = (1 - x) dx$
Integrate both sides: $\int \frac{dy}{1 + y} = \int (1 - x) dx$
This yields: $\log(1 + y) = x - \frac{x^2}{2} + C$

(A) Given differential equation: $x \cos y \,dy = (x e^x \log x + e^x) dx$
Divide both sides by $x$ (assuming $x \neq 0$):
$\cos y \,dy = \left(e^x \log x + \frac{e^x}{x}\right) dx$
Integrating both sides:
$\int \cos y \,dy = \int e^x \left(\log x + \frac{1}{x}\right) dx$
Using the standard integral formula $\int e^x \{f(x) + f'(x)\} dx = e^x f(x) + C$, where $f(x) = \log x$ and $f'(x) = \frac{1}{x}$:
$\sin y = e^x \log x + C$

(D) Given the differential equation $e^{\frac{dy}{dx}} = x+1$.
Taking the natural logarithm on both sides,we get $\frac{dy}{dx} = \log_{e}(x+1)$.
Integrating both sides with respect to $x$: $\int dy = \int \log_{e}(x+1) dx$.
Using integration by parts,let $u = \log_{e}(x+1)$ and $dv = dx$. Then $du = \frac{1}{x+1} dx$ and $v = x+1$.
$\int \log_{e}(x+1) dx = (x+1) \log_{e}(x+1) - \int \frac{x+1}{x+1} dx = (x+1) \log_{e}(x+1) - x + C$.
So,$y = (x+1) \log_{e}(x+1) - x + C$.
Given the initial condition $y(0) = 5$,substitute $x=0$ and $y=5$:
$5 = (0+1) \log_{e}(0+1) - 0 + C
\Rightarrow 5 = 1 \cdot \log_{e}(1) - 0 + C
\Rightarrow 5 = 0 - 0 + C
\Rightarrow C = 5$.
Therefore,the solution is $y = (x+1) \log_{e}(x+1) - x + 5$.

(D) Given differential equation: $(1+x) y \,dx + (1-y) x \,dy = 0$
Divide both sides by $xy$:
$\frac{1+x}{x} \,dx + \frac{1-y}{y} \,dy = 0$
$(\frac{1}{x} + 1) \,dx + (\frac{1}{y} - 1) \,dy = 0$
Integrate both sides:
$\int (\frac{1}{x} + 1) \,dx + \int (\frac{1}{y} - 1) \,dy = C_1$
$\log|x| + x + \log|y| - y = C_1$
Using the property $\log a + \log b = \log(ab)$:
$\log|xy| + x - y = C$

(D) Given the differential equation: $\frac{y}{x} \frac{dy}{dx} = \frac{1+y^2}{1+x^2}$
Separating the variables,we get: $\frac{y}{1+y^2} dy = \frac{x}{1+x^2} dx$
Integrating both sides: $\int \frac{y}{1+y^2} dy = \int \frac{x}{1+x^2} dx$
Multiply by $2$ on both sides to facilitate integration: $\int \frac{2y}{1+y^2} dy = \int \frac{2x}{1+x^2} dx$
This results in: $\ln|1+y^2| = \ln|1+x^2| + \ln C$
Using logarithmic properties: $\ln(1+y^2) = \ln(C(1+x^2))$
Thus: $1+y^2 = C(1+x^2)$
Given $x=2$ and $y=1$: $1+(1)^2 = C(1+(2)^2) \Rightarrow 2 = 5C \Rightarrow C = \frac{2}{5}$
Substituting $C$ back into the equation: $1+y^2 = \frac{2}{5}(1+x^2)$
Multiplying by $5$: $5(1+y^2) = 2(1+x^2)$

(A) Given the differential equation: $x+y \frac{dy}{dx}=\sec(x^2+y^2)$.
Let $u = x^2+y^2$. Differentiating with respect to $x$,we get $\frac{du}{dx} = 2x + 2y \frac{dy}{dx}$.
This implies $x + y \frac{dy}{dx} = \frac{1}{2} \frac{du}{dx}$.
Substituting this into the original equation,we have $\frac{1}{2} \frac{du}{dx} = \sec(u)$.
Separating the variables,we get $\frac{du}{\sec(u)} = 2 dx$,which is $\cos(u) du = 2 dx$.
Integrating both sides,we get $\int \cos(u) du = \int 2 dx$.
This results in $\sin(u) = 2x + c$.
Substituting back $u = x^2+y^2$,the general solution is $\sin(x^2+y^2) = 2x + c$.

(D) Given the differential equation: $\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$.
Let $u = x+y$. Then,differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{du}{dx}$,which implies $\frac{dy}{dx} = \frac{du}{dx} - 1$.
Substituting this into the equation:
$\frac{du}{dx} - 1 = \frac{u+1}{u-1}$
$\frac{du}{dx} = \frac{u+1}{u-1} + 1 = \frac{u+1+u-1}{u-1} = \frac{2u}{u-1}$.
Separating the variables:
$\left(\frac{u-1}{u}\right) du = 2 dx$
$(1 - \frac{1}{u}) du = 2 dx$.
Integrating both sides:
$\int (1 - \frac{1}{u}) du = \int 2 dx$
$u - \log|u| = 2x + c$.
Substituting $u = x+y$ back:
$(x+y) - \log|x+y| = 2x + c$
$y - x = \log|x+y| + c$ or $y = x + \log|x+y| + c$.

(B) Given the differential equation: $y(1+\log x)\left(\frac{dx}{dy}\right) - x \log x = 0$
Rearranging the terms,we get: $y(1+\log x) dx = x \log x dy$
Separating the variables: $\frac{(1+\log x)}{x \log x} dx = \frac{dy}{y}$
Integrating both sides: $\int \frac{1+\log x}{x \log x} dx = \int \frac{1}{y} dy$
Split the integral on the left: $\int \frac{1}{x \log x} dx + \int \frac{\log x}{x \log x} dx = \int \frac{1}{y} dy$
$\int \frac{1}{x \log x} dx + \int \frac{1}{x} dx = \int \frac{1}{y} dy$
Let $u = \log x$,then $du = \frac{1}{x} dx$. The integral becomes: $\int \frac{1}{u} du + \int \frac{1}{x} dx = \int \frac{1}{y} dy$
Integrating gives: $\log|u| + \log|x| = \log|y| + \log|c|$
$\log|\log x| + \log|x| = \log|y| + \log|c|$
Using the property $\log a + \log b = \log(ab)$: $\log|x \log x| = \log|yc|$
Taking the exponential of both sides: $x \log x = yc$