Variable separable type differential equations Questions in English

(B) Given the differential equation $\frac{dy}{dx} = \frac{2x-3y+5}{6x-9y+7}$.
Let $v = 2x-3y$. Then $\frac{dv}{dx} = 2 - 3\frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{1}{3}(2 - \frac{dv}{dx})$.
Substituting into the equation: $\frac{1}{3}(2 - \frac{dv}{dx}) = \frac{v+5}{3v+7}$.
$2 - \frac{dv}{dx} = \frac{3v+15}{3v+7} \implies \frac{dv}{dx} = 2 - \frac{3v+15}{3v+7} = \frac{6v+14-3v-15}{3v+7} = \frac{3v-1}{3v+7}$.
Separating variables: $\int \frac{3v+7}{3v-1} dv = \int dx$.
$\int (1 + \frac{8}{3v-1}) dv = \int dx \implies v + \frac{8}{3} \log |3v-1| = x + c$.
Substituting $v = 2x-3y$: $(2x-3y) + \frac{8}{3} \log |3(2x-3y)-1| = x + c$.
$x - 3y + \frac{8}{3} \log |6x-9y-1| + c = 0$.

(D) Given the differential equation: $\cos y \frac{dy}{dx} = e^x e^{\sin y} + x^2 e^{\sin y}$.
Divide both sides by $e^{\sin y}$: $e^{-\sin y} \cos y \frac{dy}{dx} = e^x + x^2$.
Let $u = \sin y$, then $\frac{du}{dx} = \cos y \frac{dy}{dx}$.
The equation becomes $e^{-u} \frac{du}{dx} = e^x + x^2$.
Integrating both sides with respect to $x$: $\int e^{-u} du = \int (e^x + x^2) dx$.
$-e^{-u} = e^x + \frac{x^3}{3} + C_1$.
Substitute $u = \sin y$ back: $-e^{-\sin y} = e^x + \frac{x^3}{3} + C_1$.
Rearranging to the form $f(x) + e^{-\sin y} = C$: $e^x + \frac{x^3}{3} + e^{-\sin y} = C$.
Comparing this with $f(x) + e^{-\sin y} = C$, we get $f(x) = e^x + \frac{x^3}{3}$.

(A) Given the differential equation: $x \frac{dy}{dx} + y = \frac{x f(xy)}{f'(xy)}$
We know that $\frac{d}{dx}(xy) = x \frac{dy}{dx} + y$.
Substituting this into the equation, we get: $\frac{d(xy)}{dx} = \frac{x f(xy)}{f'(xy)}$.
Rearranging the terms to separate the variables $xy$ and $x$: $\frac{f'(xy)}{f(xy)} d(xy) = x dx$.
Integrating both sides: $\int \frac{f'(xy)}{f(xy)} d(xy) = \int x dx$.
This yields: $\ln |f(xy)| = \frac{x^2}{2} + C$.
Taking the exponential of both sides: $|f(xy)| = e^{\frac{x^2}{2} + C} = e^C \cdot e^{x^2 / 2}$.
Letting $k = e^C$, we get: $|f(xy)| = k e^{x^2 / 2}$.

(D) Given equation: $(x+y)^{2} \frac{d y}{d x}=a^{2}, a \neq 0$
Let $x+y=t$. Then $1+\frac{d y}{d x}=\frac{d t}{d x}$, which implies $\frac{d y}{d x}=\frac{d t}{d x}-1$.
Substituting this into the given equation:
$t^{2}(\frac{d t}{d x}-1)=a^{2}$
$t^{2} \frac{d t}{d x} = a^{2}+t^{2}$
Separating the variables:
$\frac{t^{2}}{a^{2}+t^{2}} d t = d x$
Integrating both sides:
$\int \frac{t^{2}+a^{2}-a^{2}}{t^{2}+a^{2}} d t = \int d x$
$\int (1 - \frac{a^{2}}{t^{2}+a^{2}}) d t = x + C'$
$t - a^{2} \cdot \frac{1}{a} \tan^{-1}(\frac{t}{a}) = x + C'$
$t - a \tan^{-1}(\frac{t}{a}) = x + C'$
Substituting $t = x+y$ back:
$(x+y) - a \tan^{-1}(\frac{x+y}{a}) = x + C'$
$y - C' = a \tan^{-1}(\frac{x+y}{a})$
$\frac{y-C'}{a} = \tan^{-1}(\frac{x+y}{a})$
$\tan(\frac{y-C'}{a}) = \frac{x+y}{a}$
Let $C = -C'$. Then the general solution is $\tan(\frac{y+C}{a}) = \frac{x+y}{a}$.

(A) Given the differential equation: $(x+y)^{2} \frac{dy}{dx} = a^{2}$.
Let $v = x+y$. Then, 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: $v^{2} (\frac{dv}{dx} - 1) = a^{2}$.
Rearranging the terms: $v^{2} \frac{dv}{dx} = v^{2} + a^{2}$, so $\frac{dv}{dx} = \frac{v^{2} + a^{2}}{v^{2}}$.
Separating the variables: $\frac{v^{2}}{v^{2} + a^{2}} dv = dx$.
Integrating both sides: $\int \frac{v^{2}}{v^{2} + a^{2}} dv = \int dx$.
This can be written as: $\int (1 - \frac{a^{2}}{v^{2} + a^{2}}) dv = x + C'$.
Integrating gives: $v - a \tan^{-1}(\frac{v}{a}) = x + C'$.
Substituting $v = x+y$: $(x+y) - a \tan^{-1}(\frac{x+y}{a}) = x + C'$.
Simplifying: $y - a \tan^{-1}(\frac{x+y}{a}) = C'$.
Rearranging: $\frac{y-C'}{a} = \tan^{-1}(\frac{x+y}{a})$.
Taking the tangent on both sides: $\tan(\frac{y-C'}{a}) = \frac{x+y}{a}$.
Letting $-C' = C$, we get $\frac{x+y}{a} = \tan(\frac{y+C}{a})$.

(A) Given the differential equation: $\log_{e}\left(\frac{dy}{dx}\right) = x + y$.
By the definition of logarithm, we can write this as: $\frac{dy}{dx} = e^{x+y}$.
Using the property of exponents, we have: $\frac{dy}{dx} = e^x \cdot e^y$.
Separating the variables, we get: $e^{-y} dy = e^x dx$.
Integrating both sides: $\int e^{-y} dy = \int e^x dx$.
This yields: $-e^{-y} = e^x + C_1$.
Rearranging the terms, we get: $e^x + e^{-y} = -C_1$.
Letting $-C_1 = C$, the general solution is: $e^x + e^{-y} = C$.

(D) 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, we get $\int \frac{dy}{y} = \int \frac{dx}{x} + C$.
This results in $\ln|y| = \ln|x| + \ln|c|$, where $\ln|c|$ is the constant of integration.
Using logarithmic properties, $\ln|y| = \ln|cx|$, which implies $y = cx$.
The equation $y = cx$ represents a straight line passing through the origin.

(B) Given the differential equation: $\frac{dy}{dx} = e^{y+x} + e^{y-x}$.
We can rewrite the right side as: $\frac{dy}{dx} = e^y(e^x + e^{-x})$.
Separating the variables, we get: $e^{-y} dy = (e^x + e^{-x}) dx$.
Integrating both sides: $\int e^{-y} dy = \int (e^x + e^{-x}) dx$.
This yields: $-e^{-y} = e^x - e^{-x} + c$.
Multiplying by $-1$, we obtain: $e^{-y} = e^{-x} - e^x + c$.
Thus, the correct option is $B$.

(D) Given differential equation: $\frac{d y}{d x}=\frac{x+y+1}{2 x+2 y+1}$
Let $x+y=v$. Then $1+\frac{d y}{d x}=\frac{d v}{d x}$, so $\frac{d y}{d x}=\frac{d v}{d x}-1$.
Substituting this into the equation: $\frac{d v}{d x}-1=\frac{v+1}{2 v+1}$
$\frac{d v}{d x}=\frac{v+1}{2 v+1}+1 = \frac{v+1+2 v+1}{2 v+1} = \frac{3 v+2}{2 v+1}$
Separating variables: $\frac{2 v+1}{3 v+2} d v=d x$
Rewrite the numerator: $\frac{\frac{2}{3}(3 v+2)-\frac{1}{3}}{3 v+2} d v=d x$
$\left(\frac{2}{3}-\frac{1}{3(3 v+2)}\right) d v=d x$
Integrating both sides: $\int \left(\frac{2}{3}-\frac{1}{3(3 v+2)}\right) d v = \int d x + C'$
$\frac{2}{3} v - \frac{1}{9} \log |3 v+2| = x + C'$
Substitute $v=x+y$: $\frac{2}{3}(x+y) - \frac{1}{9} \log |3 x+3 y+2| = x + C'$
Multiply by $9$: $6(x+y) - \log |3 x+3 y+2| = 9 x + 9 C'$
$6 x + 6 y - 9 x - \log |3 x+3 y+2| = C$
$-3 x + 6 y - \log |3 x+3 y+2| = C$
Multiplying by $-1$: $3 x - 6 y + \log |3 x+3 y+2| = C$
Thus, the correct option is $D$.

(A) Given the differential equation: $x \frac{dy}{dx} + y = x \frac{f(xy)}{f'(xy)}$.
We know that $\frac{d}{dx}(xy) = x \frac{dy}{dx} + y$.
Substituting this into the equation, we get: $\frac{d(xy)}{dx} = x \frac{f(xy)}{f'(xy)}$.
Rearranging the terms to separate the variables, we have: $\frac{f'(xy)}{f(xy)} d(xy) = x dx$.
Integrating both sides: $\int \frac{f'(xy)}{f(xy)} d(xy) = \int x dx$.
This yields: $\ln |f(xy)| = \frac{x^2}{2} + k$, where $k$ is the constant of integration.
Taking the exponential of both sides: $|f(xy)| = e^{\frac{x^2}{2} + k} = e^k \cdot e^{\frac{x^2}{2}}$.
Letting $C = e^k$, we get: $|f(xy)| = Ce^{\frac{x^2}{2}}$.

(A) Given the differential equation: $16(\sqrt{x+9\sqrt{x}})(4+\sqrt{9+\sqrt{x}}) \cos y \, dy = (1+2 \sin y) \, dx$.
Separating the variables, we get: $\int \frac{\cos y}{1+2 \sin y} \, dy = \int \frac{dx}{16(\sqrt{x+9\sqrt{x}})(4+\sqrt{9+\sqrt{x}})}$.
Let $u = 1+2 \sin y$, then $du = 2 \cos y \, dy$, so $\int \frac{\cos y}{1+2 \sin y} \, dy = \frac{1}{2} \ln |1+2 \sin y|$.
For the $RHS$, let $t = 4+\sqrt{9+\sqrt{x}}$. Then $t-4 = \sqrt{9+\sqrt{x}}$. Squaring both sides: $(t-4)^2 = 9+\sqrt{x}$, so $\sqrt{x} = (t-4)^2 - 9$.
Differentiating $t = 4+\sqrt{9+\sqrt{x}}$, we get $dt = \frac{1}{2\sqrt{9+\sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} \, dx = \frac{dx}{4\sqrt{x(9+\sqrt{x})}} = \frac{dx}{4\sqrt{x+9\sqrt{x}}}$.
Thus, $\frac{dx}{\sqrt{x+9\sqrt{x}}} = 4 \, dt$.
The integral becomes: $\frac{1}{2} \ln |1+2 \sin y| = \int \frac{4 \, dt}{16t} = \frac{1}{4} \ln |t| + C = \frac{1}{4} \ln |4+\sqrt{9+\sqrt{x}}| + C$.
Using $y(256) = \frac{\pi}{2}$: $\frac{1}{2} \ln(1+2 \sin \frac{\pi}{2}) = \frac{1}{4} \ln(4+\sqrt{9+\sqrt{256}}) + C \implies \frac{1}{2} \ln 3 = \frac{1}{4} \ln(4+\sqrt{9+16}) + C = \frac{1}{4} \ln 9 + C = \frac{1}{2} \ln 3 + C$. Thus $C = 0$.
Now, for $y(49) = \alpha$: $\frac{1}{2} \ln(1+2 \sin \alpha) = \frac{1}{4} \ln(4+\sqrt{9+\sqrt{49}}) = \frac{1}{4} \ln(4+\sqrt{16}) = \frac{1}{4} \ln 8 = \frac{1}{4} \ln(2^3) = \frac{3}{4} \ln 2$.
So $\ln(1+2 \sin \alpha) = \frac{3}{2} \ln 2 = \ln(2^{3/2}) = \ln(2\sqrt{2})$.
Therefore, $1+2 \sin \alpha = 2\sqrt{2}$, which gives $2 \sin \alpha = 2\sqrt{2}-1$.

(A) Given the differential equation: $x dy - y dx = \sqrt{x^{2} + y^{2}} dx$.
Divide both sides by $x^{2}$ (since $x > 0$): $\frac{x dy - y dx}{x^{2}} = \frac{\sqrt{x^{2} + y^{2}}}{x^{2}} dx$.
This simplifies to: $d\left(\frac{y}{x}\right) = \sqrt{1 + \left(\frac{y}{x}\right)^{2}} \cdot \frac{1}{x} dx$.
Integrating both sides: $\int \frac{d(\frac{y}{x})}{\sqrt{1 + (\frac{y}{x})^{2}}} = \int \frac{1}{x} dx$.
Using the standard integral $\int \frac{du}{\sqrt{1 + u^{2}}} = \ln|u + \sqrt{1 + u^{2}}| + C$, we get: $\ln\left(\frac{y}{x} + \sqrt{1 + \frac{y^{2}}{x^{2}}}\right) = \ln x + C$.
Given $y(1) = 0$, substitute $x = 1, y = 0$: $\ln(0 + \sqrt{1 + 0}) = \ln(1) + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$.
Thus, $\frac{y}{x} + \sqrt{1 + \frac{y^{2}}{x^{2}}} = x$.
Multiplying by $x$: $y + \sqrt{x^{2} + y^{2}} = x^{2}$.
To find $y(3)$, substitute $x = 3$: $y + \sqrt{9 + y^{2}} = 9$.
$\sqrt{9 + y^{2}} = 9 - y$.
Squaring both sides: $9 + y^{2} = 81 - 18y + y^{2}$.
$18y = 72 \Rightarrow y = 4$.

(B) Given the differential equation $\frac{dy}{dx} = e^{x-y}$.
Using the property of exponents, we can write $\frac{dy}{dx} = \frac{e^x}{e^y}$.
By separating the variables, we get $e^y \, dy = e^x \, dx$.
Integrating both sides, we have $\int e^y \, dy = \int e^x \, dx$.
This results in $e^y = e^x + C$, where $C$ is the constant of integration.
Rearranging the terms, we get $e^x - e^y = -C$, which can be written as $e^x - e^y = c$ (where $c = -C$ is an arbitrary constant).

(B) The given differential equation is $\frac{dy}{dx} = e^{x+y} = e^x \cdot e^y$.
By separating the variables, we get $e^{-y} \, dy = e^x \, dx$.
Integrating both sides, we have $\int e^{-y} \, dy = \int e^x \, dx$.
This results in $-e^{-y} = e^x + C'$, where $C'$ is the constant of integration.
Rearranging the terms, we get $e^x + e^{-y} = -C'$.
Letting $C = -C'$, the general solution is $e^x + e^{-y} = C$.

(D) The given differential equation is $(1 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$.
Rearranging the terms to separate the variables, we get $\frac{dy}{y+1} = -\frac{\cos x}{1+\sin x} dx$.
Integrating both sides, we have $\int \frac{dy}{y+1} = -\int \frac{\cos x}{1+\sin x} dx$.
This yields $\ln|y+1| = -\ln|1+\sin x| + C$, which simplifies to $\ln|y+1| + \ln|1+\sin x| = C$.
Using the property of logarithms, we get $(y+1)(1+\sin x) = K$, where $K = e^C$.
Given the initial condition $y(0) = 0$, we substitute $x = 0$ and $y = 0$ into the equation: $(0+1)(1+\sin 0) = K$, which gives $1(1+0) = K$, so $K = 1$.
Thus, the particular solution is $(y+1)(1+\sin x) = 1$.
If the curve passes through $(\alpha, -\frac{1}{2})$, we substitute $x = \alpha$ and $y = -\frac{1}{2}$ into the equation:
$(-\frac{1}{2} + 1)(1+\sin \alpha) = 1$.
$\frac{1}{2}(1+\sin \alpha) = 1$.
$1+\sin \alpha = 2$.
$\sin \alpha = 1$.
Therefore, $\alpha = \frac{\pi}{2}$.

(C) Given the differential equation $\frac{dy}{dx} = (1+x^2)(1-y^2)$.
Separating the variables, we get $\int \frac{dy}{1-y^2} = \int (1+x^2) dx$.
Integrating both sides, we have $\frac{1}{2} \ln|\frac{1+y}{1-y}| = x + \frac{x^3}{3} + C$.
Using the initial condition $y(0) = \frac{1}{2}$, we find $C$: $\frac{1}{2} \ln|\frac{1+1/2}{1-1/2}| = 0 + 0 + C \Rightarrow C = \frac{1}{2} \ln(3)$.
Substituting $C$ back, $\frac{1}{2} \ln|\frac{1+y}{1-y}| = x + \frac{x^3}{3} + \frac{1}{2} \ln(3) \Rightarrow \ln|\frac{1+y}{1-y}| = 2(x + \frac{x^3}{3}) + \ln(3)$.
At $x=1$, $\ln|\frac{1+y(1)}{1-y(1)}| = 2(1 + \frac{1}{3}) + \ln(3) = \frac{8}{3} + \ln(3)$.
Thus, $\frac{1+y(1)}{1-y(1)} = 3e^{8/3}$.
Let $k = 3e^{8/3}$. Then $1+y(1) = k - ky(1) \Rightarrow y(1)(1+k) = k-1 \Rightarrow y(1) = \frac{k-1}{k+1}$.
$2y(1)-1 = 2(\frac{k-1}{k+1}) - 1 = \frac{2k-2-k-1}{k+1} = \frac{k-3}{k+1}$.
Substituting $k = 3e^{8/3}$, we get $\frac{3e^{8/3}-3}{3e^{8/3}+1}$. This expression simplifies to the form involving $\tan$ based on the hyperbolic identity $\tanh(u) = \frac{e^{2u}-1}{e^{2u}+1}$.

(C) Given the differential equation: $\frac{dy}{y} = (2 + \ln x) dx$.
Integrating both sides: $\int \frac{dy}{y} = \int (2 + \ln x) dx$.
$\ln y = 2x + (x \ln x - x) + C = x \ln x + x + C$.
Since the curve passes through $(1, e)$, substitute $x = 1$ and $y = e$: $\ln e = 1 \ln 1 + 1 + C$.
$1 = 0 + 1 + C$, which gives $C = 0$.
Thus, the equation of the curve is $\ln y = x \ln x + x$.
To find $f(e)$, substitute $x = e$: $\ln f(e) = e \ln e + e = e(1) + e = 2e$.
Therefore, $f(e) = e^{2e}$.

(C) The given differential equation is $\frac{dy}{dx} = (1+x^2)(1-y+y^2)$.
Separating the variables, we get $\int \frac{dy}{y^2-y+1} = \int (1+x^2) dx$.
Completing the square in the denominator: $y^2-y+1 = (y-1/2)^2 + 3/4$.
Thus, $\int \frac{dy}{(y-1/2)^2 + (\sqrt{3}/2)^2} = x + \frac{x^3}{3} + C$.
Using the formula $\int \frac{du}{u^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{u}{a})$, we get $\frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{2y-1}{\sqrt{3}}\right) = x + \frac{x^3}{3} + C$.
Given $y(0) = 1/2$, we have $\frac{2}{\sqrt{3}} \tan^{-1}(0) = 0 + 0 + C$, which implies $C = 0$.
So, $\frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{2y-1}{\sqrt{3}}\right) = x + \frac{x^3}{3}$.
At $x = 1$, $\frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{2y(1)-1}{\sqrt{3}}\right) = 1 + 1/3 = 4/3$.
$\tan^{-1} \left(\frac{2y(1)-1}{\sqrt{3}}\right) = \frac{4}{3} \cdot \frac{\sqrt{3}}{2} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$.
Therefore, $\frac{2y(1)-1}{\sqrt{3}} = \tan \left(\frac{2}{\sqrt{3}}\right)$, which gives $2y(1)-1 = \sqrt{3} \tan \left(\frac{2}{\sqrt{3}}\right)$.
Note: The provided options appear to have a calculation discrepancy in the argument of the tangent function. Based on the standard derivation, the correct value is $\sqrt{3} \tan \left(\frac{2}{\sqrt{3}}\right)$.

(C) Given the differential equation: $\frac{dy}{y} = (2 + \log_e x) dx$.
Integrating both sides: $\int \frac{dy}{y} = \int (2 + \log_e x) dx$.
$\log_e y = 2x + (x \log_e x - x) + C = x \log_e x + x + C$.
Since the curve passes through $(1, e)$, substitute $x = 1$ and $y = e$:
$\log_e e = 1 \cdot \log_e 1 + 1 + C$.
$1 = 0 + 1 + C \Rightarrow C = 0$.
Thus, the equation of the curve is $\log_e y = x \log_e x + x$.
To find $y(e)$, substitute $x = e$:
$\log_e y = e \log_e e + e = e(1) + e = 2e$.
Therefore, $y = e^{2e}$.

(D) Given the differential equation $\frac{dy}{dx} = y + 5$.
Separate the variables: $\frac{dy}{y + 5} = dx$.
Integrate both sides: $\int \frac{dy}{y + 5} = \int dx \implies \log|y + 5| = x + C$.
Using the initial condition $y(0) = 4$: $\log|4 + 5| = 0 + C \implies C = \log 9$.
Thus, $\log|y + 5| = x + \log 9$.
Rearranging gives $\log|y + 5| - \log 9 = x \implies \log|\frac{y + 5}{9}| = x$.
Exponentiating both sides: $\frac{y + 5}{9} = e^x \implies y = 9e^x - 5$.
Now, calculate $y(\log 2)$: $y(\log 2) = 9e^{\log 2} - 5$.
Since $e^{\log 2} = 2$, we get $y(\log 2) = 9(2) - 5 = 18 - 5 = 13$.

(C) Given the differential equation: $\frac{dy}{dx} = e^{x-y} + x^2e^{-y}$
Rewrite the equation as: $\frac{dy}{dx} = e^{-y}(e^x + x^2)$
Separate the variables: $e^y \ dy = (e^x + x^2) \ dx$
Integrate both sides: $\int e^y \ dy = \int (e^x + x^2) \ dx$
$e^y = e^x + \frac{x^3}{3} + c$

(B) Given equation: $e^{-x}(y + 1)dy + (\cos^2 x - \sin 2x)y dx = 0$
Divide by $y e^{-x}$: $\frac{y+1}{y} dy + e^x (\cos^2 x - \sin 2x) dx = 0$
$(1 + \frac{1}{y}) dy + e^x \cos^2 x dx - e^x \sin 2x dx = 0$
Integrate both sides: $\int (1 + \frac{1}{y}) dy + \int e^x \cos^2 x dx - \int e^x \sin 2x dx = C$
Note that $\int e^x \cos^2 x dx = \int e^x (\frac{1 + \cos 2x}{2}) dx = \frac{1}{2} e^x + \frac{1}{2} \int e^x \cos 2x dx$
Using $\int e^x \cos 2x dx = \frac{e^x}{1^2 + 2^2} (\cos 2x + 2 \sin 2x) = \frac{e^x}{5} (\cos 2x + 2 \sin 2x)$
So, $\int e^x \cos^2 x dx = \frac{e^x}{2} + \frac{e^x}{10} (\cos 2x + 2 \sin 2x) = \frac{e^x}{10} (5 + \cos 2x + 2 \sin 2x)$
Also $\int e^x \sin 2x dx = \frac{e^x}{5} (\sin 2x - 2 \cos 2x)$
Substituting back: $y + \log y + \frac{e^x}{10} (5 + \cos 2x + 2 \sin 2x - 2 \sin 2x + 4 \cos 2x) = C$
$y + \log y + \frac{e^x}{10} (5 + 5 \cos 2x) = C \implies y + \log y + e^x \frac{1 + \cos 2x}{2} = C \implies y + \log y + e^x \cos^2 x = C$
At $x = 0, y = 1$: $1 + \log 1 + e^0 \cos^2 0 = C \implies 1 + 0 + 1(1) = C \implies C = 2$
Final solution: $y + \log y + e^x \cos^2 x = 2$

(A) Let $v = x + y$. Then $\frac{dv}{dx} = 1 + \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{dv}{dx} - 1$.
Substituting into the equation: $\frac{dv}{dx} - 1 = \sin v + \cos v$.
$\frac{dv}{dx} = 1 + \sin v + \cos v$.
$\frac{dv}{1 + \sin v + \cos v} = dx$.
Using half-angle formulas $\sin v = \frac{2 \tan(v/2)}{1 + \tan^2(v/2)}$ and $\cos v = \frac{1 - \tan^2(v/2)}{1 + \tan^2(v/2)}$:
$1 + \sin v + \cos v = 1 + \frac{2 \tan(v/2) + 1 - \tan^2(v/2)}{1 + \tan^2(v/2)} = \frac{1 + \tan^2(v/2) + 2 \tan(v/2) + 1 - \tan^2(v/2)}{1 + \tan^2(v/2)} = \frac{2 + 2 \tan(v/2)}{1 + \tan^2(v/2)} = \frac{2(1 + \tan(v/2))}{\sec^2(v/2)}$.
Thus, $\int \frac{\sec^2(v/2) dv}{2(1 + \tan(v/2))} = \int dx$.
Let $u = 1 + \tan(v/2)$, then $du = \frac{1}{2} \sec^2(v/2) dv$.
The integral becomes $\int \frac{du}{u} = \int dx$, which gives $\log |u| = x + c$.
Substituting back, $\log |1 + \tan(\frac{x+y}{2})| = x + c$.

(C) Let $v = 9x + y + 5$. Then differentiating with respect to $x$, we get $\frac{dv}{dx} = 9 + \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{dv}{dx} - 9$.
Substituting into the equation: $\frac{dv}{dx} - 9 = v^2$.
$\frac{dv}{dx} = v^2 + 9$.
Separating variables: $\int \frac{dv}{v^2 + 3^2} = \int dx$.
Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + c$, we get $\frac{1}{3} \tan^{-1}(\frac{v}{3}) = x + c$.
Substituting $v$ back: $\frac{1}{3} \tan^{-1}(\frac{9x + y + 5}{3}) = x + c$.

(B) Given the slope $\frac{dy}{dx} = \frac{y - 1}{x^2 + x} = \frac{y - 1}{x(x + 1)}$.
Separating variables: $\int \frac{dy}{y - 1} = \int \frac{dx}{x(x + 1)}$.
Using partial fractions: $\frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{x + 1}$.
Integrating both sides: $\ln|y - 1| = \ln|x| - \ln|x + 1| + C = \ln|\frac{x}{x + 1}| + C$.
Since the curve passes through $(1, 0)$, substitute $x = 1, y = 0$: $\ln|0 - 1| = \ln|\frac{1}{1 + 1}| + C \implies 0 = \ln(\frac{1}{2}) + C \implies C = \ln(2)$.
Thus, $\ln|y - 1| = \ln|\frac{x}{x + 1}| + \ln(2) = \ln|\frac{2x}{x + 1}|$.
Taking exponents: $y - 1 = \frac{2x}{x + 1} \implies (y - 1)(x + 1) = 2x$.

(B) Let $x + y = v$. Differentiating with respect to $x$, we get $1 + \frac{dy}{dx} = \frac{dv}{dx}$, so $\frac{dy}{dx} = \frac{dv}{dx} - 1$.
Substituting into the equation: $\frac{dv}{dx} - 1 = \cos(v)$.
$\frac{dv}{dx} = 1 + \cos(v) = 2 \cos^2 \left( \frac{v}{2} \right)$.
Separating variables: $\frac{dv}{2 \cos^2 \left( \frac{v}{2} \right)} = dx$.
$\frac{1}{2} \sec^2 \left( \frac{v}{2} \right) dv = dx$.
Integrating both sides: $\int \frac{1}{2} \sec^2 \left( \frac{v}{2} \right) dv = \int dx$.
$\tan \left( \frac{v}{2} \right) = x + c$.
Substituting $v = x + y$ back: $\tan \left( \frac{x + y}{2} \right) = x + c$.

(D) Let $p = \frac{dy}{dx}$. The given equation is $p^2 = 6 - p$, which rearranges to $p^2 + p - 6 = 0$.
Factoring the quadratic, we get $(p + 3)(p - 2) = 0$, so $p = 2$ or $p = -3$.
Since $y = f(x)$ is a monotonically increasing function, $\frac{dy}{dx} \ge 0$, so we must have $p = 2$.
Integrating $\frac{dy}{dx} = 2$ with respect to $x$, we get $y = 2x + C$.
Using the initial condition $y(0) = 5$, we find $5 = 2(0) + C$, so $C = 5$.
Thus, the function is $y = 2x + 5$.
Evaluating at $x = 3$, we get $y(3) = 2(3) + 5 = 6 + 5 = 11$.

(A) Step $1$: Rearrange the differential equation: $x \, dy = -2y \, dx$.
Step $2$: Separate the variables: $\frac{dy}{y} = -2 \frac{dx}{x}$.
Step $3$: Integrate both sides: $\int \frac{dy}{y} = -2 \int \frac{dx}{x} \implies \ln|y| = -2 \ln|x| + C$.
Step $4$: Simplify: $\ln|y| = \ln|x^{-2}| + C \implies \ln|y| = \ln|\frac{1}{x^2}| + C \implies y = \frac{k}{x^2}$, where $k = e^C$.
Step $5$: Apply the initial condition $x = 2, y = 1$: $1 = \frac{k}{2^2} \implies 1 = \frac{k}{4} \implies k = 4$.
Step $6$: The particular solution is $y = \frac{4}{x^2}$, which simplifies to $x^2 y = 4$.

(D) Given the differential equation $\frac{dy}{dx} = y + 5$.
Separate the variables: $\frac{dy}{y + 5} = dx$.
Integrate both sides: $\int \frac{dy}{y + 5} = \int dx \implies \log|y + 5| = x + C$.
Using the initial condition $y(0) = 4$: $\log|4 + 5| = 0 + C \implies C = \log 9$.
Thus, $\log|y + 5| = x + \log 9 \implies \log|y + 5| = \log(9e^x)$.
So, $y + 5 = 9e^x \implies y = 9e^x - 5$.
Now, find $y(\log 2)$: $y(\log 2) = 9e^{\log 2} - 5 = 9(2) - 5 = 18 - 5 = 13$.

(C) Given the differential equation: $\frac{dy}{dx} = e^{x-y} + x^2e^{-y}$
Rewrite the equation as: $\frac{dy}{dx} = e^{-y}(e^x + x^2)$
Separate the variables: $e^y \ dy = (e^x + x^2) \ dx$
Integrate both sides: $\int e^y \ dy = \int (e^x + x^2) \ dx$
$e^y = e^x + \frac{x^3}{3} + c$
Thus, the correct option is $C$.

(B) Given: $e^{-x}(y + 1) dy + (\cos^2 x - \sin 2x)y dx = 0$
Divide by $y e^{-x}$: $\frac{y+1}{y} dy + \frac{\cos^2 x - \sin 2x}{e^{-x}} dx = 0$
$(1 + \frac{1}{y}) dy + e^x(\cos^2 x - 2 \sin x \cos x) dx = 0$
Integrate both sides: $\int (1 + \frac{1}{y}) dy + \int e^x \cos^2 x dx - \int e^x \sin 2x dx = C$
Using $\int e^x \cos^2 x dx = \int e^x (\frac{1 + \cos 2x}{2}) dx = \frac{1}{2} e^x + \frac{1}{2} \int e^x \cos 2x dx$
Thus, $\int e^x \cos^2 x dx - \int e^x \sin 2x dx = \frac{1}{2} e^x + \frac{1}{2} \int e^x \cos 2x dx - \int e^x \sin 2x dx$. This simplifies to $e^x \cos^2 x$.
So, $y + \log y = -e^x \cos^2 x + C$
At $x = 0, y = 1$: $1 + \log 1 = -e^0 \cos^2 0 + C \implies 1 = -1 + C \implies C = 2$
Final solution: $y + \log y + e^x \cos^2 x = 2$.

(A) Let $v = x + y$. Then $\frac{dv}{dx} = 1 + \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{dv}{dx} - 1$.
Substituting into the equation: $\frac{dv}{dx} - 1 = \sin v + \cos v$.
$\frac{dv}{dx} = 1 + \sin v + \cos v$.
Using half-angle formulas: $1 + \sin v + \cos v = 2\cos^2(\frac{v}{2}) + 2\sin(\frac{v}{2})\cos(\frac{v}{2}) = 2\cos^2(\frac{v}{2}) [1 + \tan(\frac{v}{2})]$.
So, $\int \frac{dv}{2\cos^2(\frac{v}{2}) [1 + \tan(\frac{v}{2})]} = \int dx$.
Let $u = 1 + \tan(\frac{v}{2})$, then $du = \frac{1}{2} \sec^2(\frac{v}{2}) dv = \frac{1}{2\cos^2(\frac{v}{2})} dv$.
Thus, $\int \frac{du}{u} = \int dx \implies \log |u| = x + c$.
Substituting back: $\log |1 + \tan(\frac{x + y}{2})| = x + c$.

(B) Let $v = 9x + y + 5$. Then $\frac{dv}{dx} = 9 + \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{dv}{dx} - 9$.
Substituting into the equation: $\frac{dv}{dx} - 9 = v^2 \implies \frac{dv}{dx} = v^2 + 9$.
Separating variables: $\int \frac{dv}{v^2 + 3^2} = \int dx$.
Using the integral formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + c$, we get $\frac{1}{3} \tan^{-1}(\frac{v}{3}) = x + c$.
Substituting $v = 9x + y + 5$ back: $\frac{1}{3} \tan^{-1}(\frac{9x + y + 5}{3}) = x + c$.

(B) Given the slope $\frac{dy}{dx} = \frac{y - 1}{x(x + 1)}$.
Separate the variables: $\int \frac{dy}{y - 1} = \int \frac{dx}{x(x + 1)}$.
Using partial fractions: $\frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{x + 1}$.
Integrating both sides: $\ln|y - 1| = \ln|x| - \ln|x + 1| + C = \ln|\frac{x}{x + 1}| + C$.
Since the curve passes through $(1, 0)$, substitute $x = 1, y = 0$: $\ln|0 - 1| = \ln|\frac{1}{1 + 1}| + C \implies 0 = \ln(\frac{1}{2}) + C \implies C = \ln(2)$.
Thus, $\ln|y - 1| = \ln|\frac{x}{x + 1}| + \ln(2) = \ln|\frac{2x}{x + 1}|$.
Taking the exponential: $y - 1 = \frac{2x}{x + 1} \implies (y - 1)(x + 1) = 2x$.

(B) Let $x + y = v$. Differentiating with respect to $x$, we get $1 + \frac{dy}{dx} = \frac{dv}{dx}$, so $\frac{dy}{dx} = \frac{dv}{dx} - 1$.
Substituting into the equation: $\frac{dv}{dx} - 1 = \cos v$.
$\frac{dv}{dx} = 1 + \cos v = 2 \cos^2 (\frac{v}{2})$.
Separating variables: $\frac{dv}{2 \cos^2 (\frac{v}{2})} = dx$.
$\frac{1}{2} \sec^2 (\frac{v}{2}) dv = dx$.
Integrating both sides: $\int \frac{1}{2} \sec^2 (\frac{v}{2}) dv = \int dx$.
$\tan (\frac{v}{2}) = x + c$.
Substituting $v = x + y$ back: $\tan (\frac{x + y}{2}) = x + c$.

(D) Let $u = \frac{dy}{dx}$. The given equation is $u^2 = 6 - u$, which implies $u^2 + u - 6 = 0$.
Factoring the quadratic equation, we get $(u + 3)(u - 2) = 0$.
Thus, $u = -3$ or $u = 2$.
Since $y = f(x)$ is a monotonically increasing function, $\frac{dy}{dx} > 0$, so we must have $\frac{dy}{dx} = 2$.
Integrating $\frac{dy}{dx} = 2$ with respect to $x$, we get $y = 2x + C$.
Using the initial condition $y(0) = 5$, we have $5 = 2(0) + C$, which gives $C = 5$.
Therefore, the function is $y = 2x + 5$.
To find $y(3)$, substitute $x = 3$: $y(3) = 2(3) + 5 = 6 + 5 = 11$.

(C) Given differential equation: $2x \frac{dy}{dx} - y = 3$
Rearranging the terms: $2x \frac{dy}{dx} = y + 3$
Separating the variables: $\frac{dy}{y + 3} = \frac{dx}{2x}$
Integrating both sides: $\int \frac{dy}{y + 3} = \int \frac{dx}{2x}$
$\ln|y + 3| = \frac{1}{2} \ln|x| + C$
Multiply by $2$: $2 \ln|y + 3| = \ln|x| + 2C$
$\ln(y + 3)^2 = \ln|x| + C'$
$(y + 3)^2 = kx$, where $k = e^{C'}$
This equation is of the form $(y - k_1)^2 = 4a(x - k_2)$, which represents a family of parabolas.

(A) Given differential equation: $x \, dy + 2y \, dx = 0$
Rearranging the terms: $x \, dy = -2y \, dx$
Separating variables: $\frac{dy}{2y} = -\frac{dx}{x}$
Integrating both sides: $\int \frac{1}{2y} \, dy = -\int \frac{1}{x} \, dx$
$\frac{1}{2} \ln|y| = -\ln|x| + C$
Multiply by $2$: $\ln|y| = -2\ln|x| + 2C$
$\ln|y| = \ln|x^{-2}| + C_1$ (where $C_1 = 2C$)
$y = e^{C_1} \cdot x^{-2} \implies y = \frac{K}{x^2}$ (where $K = e^{C_1}$)
Given $x = 2$ and $y = 1$: $1 = \frac{K}{2^2} \implies K = 4$
Thus, the particular solution is $y = \frac{4}{x^2}$, which simplifies to $x^2 y = 4$.

(A) Given the equation: $\int x f(x) dx + \frac{f(x)}{2} = 0$.
Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus:
$x f(x) + \frac{1}{2} f'(x) = 0$.
Rearranging the terms to separate variables:
$\frac{f'(x)}{f(x)} = -2x$.
Integrating both sides with respect to $x$:
$\int \frac{f'(x)}{f(x)} dx = \int -2x dx$.
$\ln|f(x)| = -x^2 + C_1$.
Exponentiating both sides:
$f(x) = e^{-x^2 + C_1} = e^{C_1} \cdot e^{-x^2}$.
Letting $e^{C_1} = C$, we get $f(x) = C e^{-x^2}$.