Exact differential equations Questions in English

(B) Given equation: $(x \cot y + \ln(\cos x)) dy + (\ln(\sin y) - y \tan x) dx = 0$
Rearranging the terms:
$(\ln(\sin y) dx + x \cot y dy) + (\ln(\cos x) dy - y \tan x dx) = 0$
We observe that:
$d(x \ln(\sin y)) = \ln(\sin y) dx + x \cdot \frac{1}{\sin y} \cdot \cos y dy = \ln(\sin y) dx + x \cot y dy$
And:
$d(y \ln(\cos x)) = \ln(\cos x) dy + y \cdot \frac{1}{\cos x} \cdot (-\sin x) dx = \ln(\cos x) dy - y \tan x dx$
Substituting these into the equation:
$d(x \ln(\sin y)) + d(y \ln(\cos x)) = 0$
Integrating both sides:
$x \ln(\sin y) + y \ln(\cos x) = C_1$
Using the property of logarithms $a \ln b = \ln(b^a)$ and $\ln a + \ln b = \ln(ab)$:
$\ln((\sin y)^x) + \ln((\cos x)^y) = C_1$
$\ln((\sin y)^x \cdot (\cos x)^y) = C_1$
Taking the exponential of both sides:
$(\sin y)^x \cdot (\cos x)^y = e^{C_1} = C$

(B) The given differential equation is $(x \cot y + \ln \cos x) dy + (\ln \sin y - y \tan x) dx = 0$.
Rearranging the terms,we get:
$(\ln \sin y) dx + (x \cot y) dy + (\ln \cos x) dy - (y \tan x) dx = 0$.
Group the terms as follows:
$[(\ln \sin y) dx + (x \cot y) dy] + [(\ln \cos x) dy - (y \tan x) dx] = 0$.
Notice that $d(x \ln \sin y) = (\ln \sin y) dx + x \cdot \frac{1}{\sin y} \cdot \cos y dy = (\ln \sin y) dx + x \cot y dy$.
Similarly,$d(y \ln \cos x) = (\ln \cos x) dy + y \cdot \frac{1}{\cos x} \cdot (-\sin x) dx = (\ln \cos x) dy - y \tan x dx$.
Substituting these into the equation,we get:
$d(x \ln \sin y) + d(y \ln \cos x) = 0$.
Integrating both sides,we get:
$x \ln \sin y + y \ln \cos x = c$.
Using the property of logarithms $a \ln b = \ln b^a$ and $\ln m + \ln n = \ln(mn)$,we have:
$\ln ((\sin y)^x \cdot (\cos x)^y) = c$.
Taking the exponential of both sides,we get:
$(\sin y)^x (\cos x)^y = e^c = C$ (where $C$ is a constant).
Thus,the correct option is $B$.

(C) The given differential equation is $\frac{dy}{dx} = \frac{(2+\alpha)x - \beta y + 2}{\beta x - 2\alpha y - (\beta\gamma - 4\alpha)}$.
For this to represent a circle,the equation must be of the form $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$.
Rearranging the differential equation: $(\beta x - 2\alpha y - (\beta\gamma - 4\alpha)) dy = ((2+\alpha)x - \beta y + 2) dx$.
This can be written as: $((2+\alpha)x - \beta y + 2) dx - (\beta x - 2\alpha y - (\beta\gamma - 4\alpha)) dy = 0$.
For this to be an exact differential equation representing a circle,the coefficient of $xy$ must be zero,so $\beta = 0$.
Substituting $\beta = 0$ into the equation: $((2+\alpha)x + 2) dx - (-2\alpha y + 4\alpha) dy = 0$.
Integrating both sides: $\int ((2+\alpha)x + 2) dx = \int (-2\alpha y + 4\alpha) dy$.
$\frac{(2+\alpha)x^2}{2} + 2x = -\alpha y^2 + 4\alpha y + C$.
Since it passes through the origin $(0,0)$,$C = 0$.
Rearranging: $\frac{2+\alpha}{2} x^2 + \alpha y^2 + 2x - 4\alpha y = 0$.
For this to be a circle,the coefficients of $x^2$ and $y^2$ must be equal: $\frac{2+\alpha}{2} = \alpha \Rightarrow 2+\alpha = 2\alpha \Rightarrow \alpha = 2$.
Substituting $\alpha = 2$: $\frac{4}{2} x^2 + 2y^2 + 2x - 8y = 0 \Rightarrow 2x^2 + 2y^2 + 2x - 8y = 0$.
Dividing by $2$: $x^2 + y^2 + x - 4y = 0$.
The center is $(-\frac{1}{2}, 2)$ and the radius is $r = \sqrt{(-\frac{1}{2})^2 + (2)^2} = \sqrt{\frac{1}{4} + 4} = \sqrt{\frac{17}{4}} = \frac{\sqrt{17}}{2}$.

(A) Given differential equation is $(6x^2 - 2xy - 18x + 3y)dx - (x^2 - 3x)dy = 0$.
Rearranging the terms: $(6x^2 - 18x)dx + (3y - 2xy)dx - (x^2 - 3x)dy = 0$.
We can write this as: $(6x^2 - 18x)dx + 3ydx - 2xydx - x^2dy + 3xdy = 0$.
Grouping terms: $(6x^2 - 18x)dx + 3(ydx + xdy) - (2xydx + x^2dy) = 0$.
Integrating each part: $\int (6x^2 - 18x)dx + 3\int d(xy) - \int d(x^2y) = 0$.
This simplifies to: $2x^3 - 9x^2 + 3xy - x^2y + c = 0$.