Homogeneous differential equations Questions in English

(B) function $f(x, y)$ is said to be homogeneous of degree $n$ if $f(tx, ty) = t^n f(x, y)$.
Given $f(x, y) = \frac{1}{x + y}$.
Substitute $x$ with $tx$ and $y$ with $ty$:
$f(tx, ty) = \frac{1}{tx + ty} = \frac{1}{t(x + y)} = t^{-1} \cdot \frac{1}{x + y} = t^{-1} f(x, y)$.
Comparing this with $f(tx, ty) = t^n f(x, y)$,we get $n = -1$.
Therefore,the function is homogeneous of degree $-1$.

(A) The given equation is a homogeneous differential equation: $\frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2} = 1 + \frac{y}{x} + \left( \frac{y}{x} \right)^2$.
Substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = 1 + v + v^2$.
Simplifying,we get $x \frac{dv}{dx} = 1 + v^2$.
Separating the variables: $\frac{dv}{1 + v^2} = \frac{dx}{x}$.
Integrating both sides: $\int \frac{dv}{1 + v^2} = \int \frac{dx}{x}$.
This yields $\tan^{-1}(v) = \log |x| + c$.
Substituting $v = \frac{y}{x}$ back,we get $\tan^{-1} \left( \frac{y}{x} \right) = \log |x| + c$.

(D) The given differential equation is $2xy \frac{dy}{dx} = x^2 + 3y^2$,which is a homogeneous differential equation.
Rewrite it as $\frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy}$.
Substitute $y = vx$,so $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{x^2 + 3(vx)^2}{2x(vx)} = \frac{x^2(1 + 3v^2)}{2x^2v} = \frac{1 + 3v^2}{2v}$.
Then,$x \frac{dv}{dx} = \frac{1 + 3v^2}{2v} - v = \frac{1 + 3v^2 - 2v^2}{2v} = \frac{1 + v^2}{2v}$.
Separating variables: $\frac{2v}{1 + v^2} dv = \frac{1}{x} dx$.
Integrating both sides: $\int \frac{2v}{1 + v^2} dv = \int \frac{1}{x} dx$.
This gives $\ln(1 + v^2) = \ln|x| + \ln|p|$,where $\ln|p|$ is the constant of integration.
So,$1 + v^2 = px$,which means $1 + \frac{y^2}{x^2} = px$.
Multiplying by $x^2$,we get $x^2 + y^2 = px^3$.

(B) Given differential equation is $(x^2 + y^2)dx = 2xydy$.
This can be rewritten as $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{x^2 + v^2x^2}{2x(vx)} = \frac{x^2(1 + v^2)}{2x^2v} = \frac{1 + v^2}{2v}$.
$x\frac{dv}{dx} = \frac{1 + v^2}{2v} - v = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v}$.
Separating the variables: $\frac{2v}{1 - v^2} dv = \frac{dx}{x}$.
Integrating both sides: $-\ln|1 - v^2| = \ln|x| + \ln|c|$.
$-\ln|1 - v^2| = \ln|cx| \implies \ln|1 - v^2|^{-1} = \ln|cx| \implies \frac{1}{1 - v^2} = cx$.
Since $v = \frac{y}{x}$,we have $\frac{1}{1 - (y/x)^2} = cx \implies \frac{x^2}{x^2 - y^2} = cx \implies x = c(x^2 - y^2)$.

(B) Given the homogeneous differential equation $\frac{dy}{dx} = \frac{x + y}{x - y}$.
Substitute $y = vx$,which implies $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{x + vx}{x - vx} = \frac{1 + v}{1 - v}$.
Rearranging terms: $x\frac{dv}{dx} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v + v^2}{1 - v} = \frac{1 + v^2}{1 - v}$.
Separating variables: $\frac{dx}{x} = \frac{1 - v}{1 + v^2} dv = \left( \frac{1}{1 + v^2} - \frac{v}{1 + v^2} \right) dv$.
Integrating both sides: $\int \frac{dx}{x} = \int \frac{1}{1 + v^2} dv - \int \frac{v}{1 + v^2} dv$.
$\ln|x| = \tan^{-1}(v) - \frac{1}{2} \ln(1 + v^2) + C$.
Substitute $v = \frac{y}{x}$: $\ln|x| = \tan^{-1}(\frac{y}{x}) - \frac{1}{2} \ln(1 + \frac{y^2}{x^2}) + \ln|c|$.
$\ln|x| = \tan^{-1}(\frac{y}{x}) - \frac{1}{2} \ln(\frac{x^2 + y^2}{x^2}) + \ln|c|$.
$\ln|x| = \tan^{-1}(\frac{y}{x}) - \frac{1}{2} [\ln(x^2 + y^2) - \ln(x^2)] + \ln|c|$.
$\ln|x| = \tan^{-1}(\frac{y}{x}) - \frac{1}{2} \ln(x^2 + y^2) + \ln|x| + \ln|c|$.
$0 = \tan^{-1}(\frac{y}{x}) - \ln((x^2 + y^2)^{1/2}) + \ln|c|$.
$\ln((x^2 + y^2)^{1/2}) = \tan^{-1}(\frac{y}{x}) + \ln|c|$.
Taking the exponential of both sides: $(x^2 + y^2)^{1/2} = c e^{\tan^{-1}(y/x)}$ or $c(x^2 + y^2)^{1/2} = e^{\tan^{-1}(y/x)}$.

(B) Given differential equation: $x \, dy - y \, dx = \sqrt{x^2 + y^2} \, dx$
Divide by $dx$: $x \frac{dy}{dx} - y = \sqrt{x^2 + y^2}$
$x \frac{dy}{dx} = y + \sqrt{x^2 + y^2}$
$\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}$
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation:
$v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2 + v^2x^2}}{x} = v + \sqrt{1 + v^2}$
$x \frac{dv}{dx} = \sqrt{1 + v^2}$
Separating variables: $\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}$
Integrating both sides: $\ln|v + \sqrt{1 + v^2}| = \ln|x| + \ln|c|$
$v + \sqrt{1 + v^2} = cx$
Substitute $v = \frac{y}{x}$:
$\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = cx$
$\frac{y + \sqrt{x^2 + y^2}}{x} = cx$
$y + \sqrt{x^2 + y^2} = cx^2$

(C) Given differential equation is $(x + y)dx + xdy = 0$.
Rearranging the terms,we get $xdy = -(x + y)dx$,which implies $\frac{dy}{dx} = -\frac{x + y}{x} = -(1 + \frac{y}{x})$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation: $v + x\frac{dv}{dx} = -(1 + v) = -1 - v$.
$x\frac{dv}{dx} = -1 - 2v$.
Separating the variables: $\frac{dv}{1 + 2v} = -\frac{dx}{x}$.
Integrating both sides: $\int \frac{dv}{1 + 2v} = -\int \frac{dx}{x}$.
$\frac{1}{2} \ln|1 + 2v| = -\ln|x| + C_1$.
$\ln|1 + 2v| = -2\ln|x| + 2C_1 = \ln|x^{-2}| + \ln|C|$.
$1 + 2v = \frac{C}{x^2}$.
Substituting $v = \frac{y}{x}$: $1 + 2(\frac{y}{x}) = \frac{C}{x^2}$.
$\frac{x + 2y}{x} = \frac{C}{x^2} \implies x(x + 2y) = C \implies x^2 + 2xy = C$.

(B) Given the differential equation $x + y\frac{dy}{dx} = 2y$.
Rearranging the terms,we get $y\frac{dy}{dx} = 2y - x$,which implies $\frac{dy}{dx} = \frac{2y - x}{y}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{2vx - x}{vx} = \frac{2v - 1}{v}$.
$x\frac{dv}{dx} = \frac{2v - 1}{v} - v = \frac{2v - 1 - v^2}{v} = -\frac{(v - 1)^2}{v}$.
Separating the variables: $\frac{v}{(v - 1)^2} dv = -\frac{dx}{x}$.
Using partial fractions: $\frac{v}{(v - 1)^2} = \frac{v - 1 + 1}{(v - 1)^2} = \frac{1}{v - 1} + \frac{1}{(v - 1)^2}$.
Integrating both sides: $\int \left( \frac{1}{v - 1} + \frac{1}{(v - 1)^2} \right) dv = -\int \frac{dx}{x}$.
$\log |v - 1| - \frac{1}{v - 1} = -\log |x| + c$.
Since $v = \frac{y}{x}$,we have $v - 1 = \frac{y - x}{x}$.
$\log |\frac{y - x}{x}| + \log |x| = \frac{1}{\frac{y - x}{x}} + c$.
$\log |y - x| = \frac{x}{y - x} + c$.

(A) Given the differential equation $\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$.
Since this is a homogeneous differential equation,we substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{x(vx)}{x^2 + (vx)^2} = \frac{vx^2}{x^2(1 + v^2)} = \frac{v}{1 + v^2}$.
Now,$x \frac{dv}{dx} = \frac{v}{1 + v^2} - v = \frac{v - v - v^3}{1 + v^2} = -\frac{v^3}{1 + v^2}$.
Separating the variables: $\frac{1 + v^2}{v^3} dv = -\frac{dx}{x}$,which simplifies to $(\frac{1}{v^3} + \frac{1}{v}) dv = -\frac{dx}{x}$.
Integrating both sides: $\int (v^{-3} + v^{-1}) dv = -\int \frac{1}{x} dx$.
This gives $-\frac{1}{2v^2} + \ln|v| = -\ln|x| + C$.
Substituting $v = \frac{y}{x}$: $-\frac{x^2}{2y^2} + \ln(\frac{y}{x}) = -\ln|x| + C$.
$-\frac{x^2}{2y^2} + \ln|y| - \ln|x| = -\ln|x| + C$,which simplifies to $\ln|y| - \frac{x^2}{2y^2} = C$.
Rearranging gives $\ln|y| = \frac{x^2}{2y^2} + C$,so $y = e^{\frac{x^2}{2y^2} + C} = k e^{\frac{x^2}{2y^2}}$.
Squaring both sides: $y^2 = k^2 e^{\frac{x^2}{y^2}}$. Letting $a = k^2$,we get $ay^2 = e^{x^2/y^2}$.

(A) Given the differential equation $\frac{dy}{dx} = \frac{x}{2y - x}$.
Since this is a homogeneous differential equation,we substitute $y = vx$,which implies $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{x}{2vx - x} = \frac{1}{2v - 1}$.
Rearranging for $\frac{dv}{dx}$: $x\frac{dv}{dx} = \frac{1}{2v - 1} - v = \frac{1 - v(2v - 1)}{2v - 1} = \frac{1 - 2v^2 + v}{2v - 1} = -\frac{2v^2 - v - 1}{2v - 1} = -\frac{(2v + 1)(v - 1)}{2v - 1}$.
Separating the variables: $\frac{2v - 1}{(2v + 1)(v - 1)} dv = -\frac{dx}{x}$.
Using partial fractions: $\frac{2v - 1}{(2v + 1)(v - 1)} = \frac{A}{2v + 1} + \frac{B}{v - 1}$.
Solving for coefficients,we get $\frac{4/3}{2v + 1} + \frac{1/3}{v - 1} dv = -\frac{dx}{x}$.
Integrating both sides: $\frac{2}{3}\ln|2v + 1| + \frac{1}{3}\ln|v - 1| = -\ln|x| + C$.
Multiplying by $3$: $2\ln|2v + 1| + \ln|v - 1| = -3\ln|x| + C'$.
$\ln|(2v + 1)^2(v - 1)| = \ln|x^{-3}| + C'$.
$(2v + 1)^2(v - 1) = \frac{c}{x^3}$.
Substituting $v = \frac{y}{x}$: $(2\frac{y}{x} + 1)^2(\frac{y}{x} - 1) = \frac{c}{x^3}$.
$(\frac{2y + x}{x})^2(\frac{y - x}{x}) = \frac{c}{x^3}$.
$(2y + x)^2(y - x) = c$. Since $(y - x) = -(x - y)$,we get $(x - y)(x + 2y)^2 = c$.

(A) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} \left( \log \frac{y}{x} + 1 \right)$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation:
$v + x \frac{dv}{dx} = v (\log v + 1)$
$v + x \frac{dv}{dx} = v \log v + v$
$x \frac{dv}{dx} = v \log v$
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 $du = \frac{1}{v} dv$. The integral becomes:
$\int \frac{du}{u} = \int \frac{dx}{x}$
$\log |u| = \log |x| + \log |c|$
$\log |\log v| = \log |xc|$
$\log v = xc$
Substituting $v = \frac{y}{x}$ back:
$\log \left( \frac{y}{x} \right) = cx$.

(A) Given the homogeneous differential equation $\frac{dy}{dx} = \frac{y - x}{y + x}$.
Substitute $y = vx$,so $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting into the equation: $v + x\frac{dv}{dx} = \frac{vx - x}{vx + x} = \frac{v - 1}{v + 1}$.
Rearranging: $x\frac{dv}{dx} = \frac{v - 1}{v + 1} - v = \frac{v - 1 - v^2 - v}{v + 1} = -\frac{v^2 + 1}{v + 1}$.
Separating variables: $\int \frac{dx}{x} = -\int \frac{v + 1}{v^2 + 1} dv$.
Integrating both sides: $\log_e x = -\left[ \frac{1}{2} \int \frac{2v}{v^2 + 1} dv + \int \frac{1}{v^2 + 1} dv \right] + c$.
$\log_e x = -\frac{1}{2} \log_e(v^2 + 1) - \tan^{-1}v + c$.
Multiply by $-2$: $-2\log_e x = \log_e(v^2 + 1) + 2\tan^{-1}v + c$.
Substitute $v = \frac{y}{x}$: $-2\log_e x = \log_e\left(\frac{y^2}{x^2} + 1\right) + 2\tan^{-1}\left(\frac{y}{x}\right) + c$.
$-2\log_e x = \log_e\left(\frac{y^2 + x^2}{x^2}\right) + 2\tan^{-1}\left(\frac{y}{x}\right) + c$.
$-2\log_e x = \log_e(x^2 + y^2) - 2\log_e x + 2\tan^{-1}\left(\frac{y}{x}\right) + c$.
Thus,$\log_e(x^2 + y^2) + 2\tan^{-1}\left(\frac{y}{x}\right) + c = 0$.

(A) Given the homogeneous differential equation $\frac{dy}{dx} = \frac{x - y}{x + y}$.
Substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting into the equation: $v + x \frac{dv}{dx} = \frac{x - vx}{x + vx} = \frac{1 - v}{1 + v}$.
Rearranging terms: $x \frac{dv}{dx} = \frac{1 - v}{1 + v} - v = \frac{1 - v - v - v^2}{1 + v} = \frac{1 - 2v - v^2}{1 + v}$.
Separating variables: $\frac{1 + v}{1 - 2v - v^2} dv = \frac{dx}{x}$.
Multiply by $-1/2$: $-\frac{1}{2} \frac{-2 - 2v}{1 - 2v - v^2} dv = \frac{dx}{x}$.
Integrating both sides: $-\frac{1}{2} \ln|1 - 2v - v^2| = \ln|x| + C_1$.
Multiply by $-2$: $\ln|1 - 2v - v^2| = -2 \ln|x| + C_2 = \ln|x^{-2}| + C_2$.
Thus,$1 - 2v - v^2 = \frac{C}{x^2}$.
Substituting $v = \frac{y}{x}$: $1 - 2(\frac{y}{x}) - (\frac{y}{x})^2 = \frac{C}{x^2}$.
Multiplying by $x^2$: $x^2 - 2xy - y^2 = C$,which can be rewritten as $y^2 + 2xy - x^2 = c$.

(B) Given differential equation: $(2x - y + 1)dx + (2y - x + 1)dy = 0$.
Rearranging,we get $\frac{dy}{dx} = -\frac{2x - y + 1}{2y - x + 1} = \frac{2x - y + 1}{x - 2y - 1}$.
Let $x = X + h$ and $y = Y + k$. Substituting these,we get $\frac{dY}{dX} = \frac{2X - Y + 2h - k + 1}{X - 2Y + h - 2k - 1}$.
For the equation to be homogeneous,we set $2h - k + 1 = 0$ and $h - 2k - 1 = 0$. Solving these gives $h = -1$ and $k = -1$.
Thus,$\frac{dY}{dX} = \frac{2X - Y}{X - 2Y}$.
Substitute $Y = vX$,so $\frac{dY}{dX} = v + X\frac{dv}{dX}$.
$v + X\frac{dv}{dX} = \frac{2 - v}{1 - 2v} \implies X\frac{dv}{dX} = \frac{2 - v}{1 - 2v} - v = \frac{2 - v - v + 2v^2}{1 - 2v} = \frac{2(v^2 - v + 1)}{1 - 2v}$.
Separating variables: $\frac{dX}{X} = \frac{1 - 2v}{2(v^2 - v + 1)} dv$.
Integrating both sides: $\int \frac{dX}{X} = -\frac{1}{2} \int \frac{2v - 1}{v^2 - v + 1} dv$.
$\ln|X| = -\frac{1}{2} \ln|v^2 - v + 1| + \ln|c_1| \implies \ln|X| + \frac{1}{2} \ln|v^2 - v + 1| = \ln|c_1|$.
$X^2(v^2 - v + 1) = c$. Substituting $v = \frac{Y}{X}$,we get $X^2(\frac{Y^2}{X^2} - \frac{Y}{X} + 1) = c \implies Y^2 - XY + X^2 = c$.
Substituting $X = x + 1$ and $Y = y + 1$: $(y + 1)^2 - (y + 1)(x + 1) + (x + 1)^2 = c$.
Expanding: $(y^2 + 2y + 1) - (xy + x + y + 1) + (x^2 + 2x + 1) = c$.
$x^2 + y^2 - xy + x + y = c'$ (where $c' = c - 1$).

(A) Given the differential equation $(x^2 + y^2)dy = xy dx$.
Rearranging the terms,we get $x^2 dy + y^2 dy = xy dx$.
$x^2 dy - xy dx = -y^2 dy$.
Dividing by $xy^2$,we get $\frac{x dy - y dx}{y^2} = -\frac{dy}{x}$.
This is equivalent to $d(\frac{x}{y}) = -\frac{dy}{x}$ (Note: The standard form is $x dy - y dx = x^2 d(\frac{y}{x})$ or $y dx - x dy = y^2 d(\frac{x}{y})$).
Let us rewrite: $x^2 dy = xy dx - y^2 dy = y(x dx - y dy)$ is not correct. Let's use substitution $y = vx$,$dy = v dx + x dv$.
$(x^2 + v^2 x^2)(v dx + x dv) = x(vx) dx$.
$x^2(1 + v^2)(v dx + x dv) = vx^2 dx$.
$(1 + v^2)(v dx + x dv) = v dx$.
$v dx + x dv + v^3 dx + v^2 x dv = v dx$.
$x dv(1 + v^2) = -v^3 dx$.
$\frac{1 + v^2}{v^3} dv = -\frac{dx}{x}$.
Integrating both sides: $\int (v^{-3} + v^{-1}) dv = -\int \frac{dx}{x}$.
$-\frac{1}{2v^2} + \ln|v| = -\ln|x| + C$.
$-\frac{x^2}{2y^2} + \ln|\frac{y}{x}| = -\ln|x| + C$.
$-\frac{x^2}{2y^2} + \ln|y| - \ln|x| = -\ln|x| + C$.
$-\frac{x^2}{2y^2} + \ln|y| = C$.
Given $y(1) = 1$,so $-\frac{1^2}{2(1)^2} + \ln(1) = C \implies C = -\frac{1}{2}$.
So,$-\frac{x^2}{2y^2} + \ln|y| = -\frac{1}{2}$.
For $y(x_0) = e$,$-\frac{x_0^2}{2e^2} + \ln(e) = -\frac{1}{2}$.
$-\frac{x_0^2}{2e^2} + 1 = -\frac{1}{2}$.
$-\frac{x_0^2}{2e^2} = -\frac{3}{2}$.
$x_0^2 = 3e^2 \implies x_0 = \sqrt{3}e$.

(C) The slope of the integral curve at any point $(x, y)$ is given by the differential equation $\frac{dy}{dx} = \frac{x^2 + y^2}{x^2 - y^2}$.
To find the slope at the point $(1, 0)$,we substitute $x = 1$ and $y = 0$ into the given differential equation.
$\left( \frac{dy}{dx} \right)_{(1, 0)} = \frac{1^2 + 0^2}{1^2 - 0^2} = \frac{1 + 0}{1 - 0} = \frac{1}{1} = 1$.
Therefore,the slope at the point $(1, 0)$ is $1$.

(A) Given the differential equation: $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{x^2 + v^2x^2}{2vx^2} = \frac{1 + v^2}{2v}$.
$x\frac{dv}{dx} = \frac{1 + v^2}{2v} - v = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v}$.
Separating variables: $\frac{2v}{1 - v^2} dv = \frac{dx}{x}$.
Integrating both sides: $-\ln|1 - v^2| = \ln|x| + \ln|c|$.
$-\ln|1 - \frac{y^2}{x^2}| = \ln|xc| \implies \ln|\frac{x^2}{x^2 - y^2}| = \ln|xc|$.
$\frac{x^2}{x^2 - y^2} = xc \implies x = c(x^2 - y^2)$.
Since the curve passes through $(2, 1)$,we have $2 = c(2^2 - 1^2) = c(3) \implies c = \frac{2}{3}$.
Thus,$x = \frac{2}{3}(x^2 - y^2)$,which simplifies to $3x = 2(x^2 - y^2)$.

(A) Given the differential equation $x \frac{dy}{dx} = y(\log y - \log x + 1)$.
Dividing by $x$,we get $\frac{dy}{dx} = \frac{y}{x} (\log(\frac{y}{x}) + 1)$.
This is a homogeneous differential equation.
Let $v = \frac{y}{x}$,so $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = v(\log v + 1)$.
$v + x \frac{dv}{dx} = v \log v + v$.
$x \frac{dv}{dx} = v \log v$.
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 $du = \frac{1}{v} dv$. The integral becomes $\int \frac{du}{u} = \log x + C$.
$\log(\log v) = \log x + \log c = \log(cx)$.
Taking the exponential of both sides: $\log v = cx$.
Since $v = \frac{y}{x}$,we have $\log(\frac{y}{x}) = cx$.
Therefore,$\frac{y}{x} = e^{cx}$,which implies $y = x e^{cx}$.

(A) Given the differential equation: ${y^2}\,dx + ({x^2} - xy + {y^2})\,dy = 0$
Rearranging the terms,we get: $\frac{{dx}}{{dy}} = -\frac{{{x^2} - xy + {y^2}}}{{{y^2}}}$
This is a homogeneous differential equation. Let $x = vy$,then $\frac{{dx}}{{dy}} = v + y\frac{{dv}}{{dy}}$.
Substituting these into the equation: $v + y\frac{{dv}}{{dy}} = -\left( {\frac{{{v^2}{y^2} - vy^2 + {y^2}}}{{{y^2}}}} \right)$
$v + y\frac{{dv}}{{dy}} = -({v^2} - v + 1)$
$y\frac{{dv}}{{dy}} = -{v^2} + v - 1 - v = -({v^2} + 1)$
Separating the variables: $\frac{{dv}}{{{v^2} + 1}} = -\frac{{dy}}{y}$
Integrating both sides: $\int \frac{{dv}}{{{v^2} + 1}} = -\int \frac{{dy}}{y}$
${\tan ^{ - 1}}(v) = -\log |y| + C$
Substituting $v = \frac{x}{y}$ back: ${\tan ^{ - 1}}\left( {\frac{x}{y}} \right) + \log |y| = C$
Thus,the general solution is ${\tan ^{ - 1}}\left( {\frac{x}{y}} \right) + \log y + c = 0$.

(C) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} - \cos^2\left( \frac{y}{x} \right)$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation: $v + x\frac{dv}{dx} = v - \cos^2(v)$.
This simplifies to $x\frac{dv}{dx} = -\cos^2(v)$,or $\sec^2(v) dv = -\frac{dx}{x}$.
Integrating both sides: $\int \sec^2(v) dv = -\int \frac{1}{x} dx$.
$\tan(v) = -\log|x| + C$.
Substituting $v = \frac{y}{x}$: $\tan\left( \frac{y}{x} \right) = -\log|x| + C$.
The curve passes through $\left( 1, \frac{\pi}{4} \right)$,so $\tan\left( \frac{\pi/4}{1} \right) = -\log(1) + C \implies 1 = 0 + C \implies C = 1$.
Thus,$\tan\left( \frac{y}{x} \right) = -\log(x) + 1 = \log(e) - \log(x) = \log\left( \frac{e}{x} \right)$.
Therefore,$y = x\tan^{-1}\left[ \log\left( \frac{e}{x} \right) \right]$.

(C) When three fair coins are tossed,the sample space is $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
Since the condition is that both heads and tails appear,we exclude the cases $HHH$ and $TTT$.
Thus,the reduced sample space $S'$ is $\{HHT, HTH, THH, HTT, THT, TTH\}$,so $n(S') = 6$.
Let $E$ be the event that exactly one head appears. The outcomes in $E$ are $\{HTT, THT, TTH\}$,so $n(E) = 3$.
The required probability is $P(E) = \frac{n(E)}{n(S')} = \frac{3}{6} = \frac{1}{2}$.

(C) Let $T$ denote a tail and $H$ denote a head.
Given that a head does not occur on the first two tosses,the outcomes for the first two tosses are $(T, T)$.
The experiment continues until a head appears or the coin is tossed $5$ times.
For the coin to be tossed $5$ times,we must not get a head on the $3^{rd}$ and $4^{th}$ tosses.
The probability of getting a tail on the $3^{rd}$ toss is $P(T_3) = \frac{1}{2}$.
The probability of getting a tail on the $4^{th}$ toss is $P(T_4) = \frac{1}{2}$.
Since these are independent events,the probability that the coin is tossed $5$ times given the first two were tails is $P(T_3) \times P(T_4) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

(B) Let $A$ be the event that the sum of the numbers is greater than $7$.
Let $B$ be the event that $4$ appears on the first die.
The sample space for the first die showing $4$ is $S = \{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)\}$.
The total number of outcomes in $B$ is $n(B) = 6$.
The outcomes where the sum is greater than $7$ are $(4, 4), (4, 5), (4, 6)$.
Thus,the number of favorable outcomes is $n(A \cap B) = 3$.
The conditional probability is $P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{3}{6} = \frac{1}{2}$.

(D) Total number of balls = $3 + 6 + 7 = 16$.
Let $A$ be the event that the first ball is white.
Let $B$ be the event that the second ball is blue.
Since the first ball is not replaced,the events are dependent.
The probability of drawing a white ball first is $P(A) = \frac{6}{16} = \frac{3}{8}$.
After drawing one white ball,the remaining number of balls in the bag is $15$.
The number of blue balls remains $7$.
The conditional probability of drawing a blue ball given that the first was white is $P(B|A) = \frac{7}{15}$.
The probability that the first ball is white and the second is blue is $P(A \cap B) = P(A) \times P(B|A) = \frac{6}{16} \times \frac{7}{15} = \frac{3}{8} \times \frac{7}{15} = \frac{1}{8} \times \frac{7}{5} = \frac{7}{40}$.

(B) Let $A$ be the event that the number $4$ appears at least once,and $B$ be the event that the sum of the numbers is $6$.
The sample space for the sum being $6$ is $B = \{(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)\}$.
Thus,the total number of outcomes in $B$ is $n(B) = 5$.
The event $A \cap B$ represents the outcomes where the sum is $6$ $AND$ the number $4$ appears at least once.
Looking at the set $B$,the outcomes containing $4$ are $(2, 4)$ and $(4, 2)$.
So,$A \cap B = \{(2, 4), (4, 2)\}$,which means $n(A \cap B) = 2$.
The conditional probability is given by $P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{2}{5}$.

(A) Let $P(\overline{B}) = x$. Then $P(B) = 1 - x$.
Since $A$ and $B$ are independent events,$P(A \cap B) = P(A)P(B) = \frac{1}{5}(1 - x)$.
We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $\frac{7}{10} = \frac{1}{5} + (1 - x) - \frac{1}{5}(1 - x)$.
$\frac{7}{10} = \frac{1}{5} + (1 - x)(1 - \frac{1}{5})$.
$\frac{7}{10} - \frac{2}{10} = (1 - x)(\frac{4}{5})$.
$\frac{5}{10} = (1 - x)(\frac{4}{5})$.
$\frac{1}{2} = (1 - x)(\frac{4}{5})$.
$1 - x = \frac{1}{2} \times \frac{5}{4} = \frac{5}{8}$.
Since $P(\overline{B}) = x$,we have $x = 1 - \frac{5}{8} = \frac{3}{8}$.

(D) Given the differential equation $\frac{dy}{dx} = \frac{x + y}{x} = 1 + \frac{y}{x}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation,we get $v + x \frac{dv}{dx} = 1 + v$.
Subtracting $v$ from both sides,we have $x \frac{dv}{dx} = 1$.
Separating the variables,we get $dv = \frac{dx}{x}$.
Integrating both sides,we get $v = \ln|x| + C$.
Since $y = vx$,we have $\frac{y}{x} = \ln|x| + C$,which implies $y = x \ln|x| + Cx$.
Given the condition $y(1) = 1$,we substitute $x = 1$ and $y = 1$:
$1 = 1 \cdot \ln(1) + C(1) \Rightarrow 1 = 0 + C \Rightarrow C = 1$.
Thus,the solution is $y = x \ln x + x$.

(C) Given the differential equation: $x\frac{dy}{dx} = y - x\tan \left( \frac{y}{x} \right)$.
Dividing by $x$,we get: $\frac{dy}{dx} = \frac{y}{x} - \tan \left( \frac{y}{x} \right)$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation: $v + x\frac{dv}{dx} = v - \tan v$.
Simplifying: $x\frac{dv}{dx} = -\tan v$.
Separating the variables: $\frac{dv}{\tan v} = -\frac{dx}{x}$,which is $\cot v \, dv = -\frac{dx}{x}$.
Integrating both sides: $\int \cot v \, dv = -\int \frac{dx}{x}$.
$\ln |\sin v| = -\ln |x| + \ln |c|$.
$\ln |\sin v| + \ln |x| = \ln |c|$.
$\ln |x \sin v| = \ln |c|$.
$x \sin v = c$.
Substituting $v = \frac{y}{x}$ back: $x \sin \left( \frac{y}{x} \right) = c$.

(C) Given the differential equation $\frac{dy}{dx} = \frac{y}{x} - \sin^2\left( \frac{y}{x} \right)$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting into the equation: $v + x \frac{dv}{dx} = v - \sin^2 v$.
This simplifies to $x \frac{dv}{dx} = -\sin^2 v$,or $-\csc^2 v \, dv = \frac{dx}{x}$.
Integrating both sides: $-\int \csc^2 v \, dv = \int \frac{dx}{x}$.
This gives $\cot v = \log_e x + C$,so $\cot\left( \frac{y}{x} \right) = \log_e x + C$.
The curve passes through $\left( 1, \frac{\pi}{4} \right)$,so $\cot\left( \frac{\pi/4}{1} \right) = \log_e 1 + C$.
Since $\cot(\pi/4) = 1$ and $\log_e 1 = 0$,we get $C = 1$.
Thus,$\cot\left( \frac{y}{x} \right) = \log_e x + 1 = \log_e x + \log_e e = \log_e(ex)$.
Therefore,$\frac{y}{x} = \cot^{-1}(\log_e ex)$,which implies $y = x \cot^{-1}(\log_e ex)$.

(D) function $f(x, y)$ is homogeneous of degree $n$ if $f(\lambda x, \lambda y) = \lambda^n f(x, y)$.
$(A)$ $f(\lambda x, \lambda y) = \frac{\lambda x - \lambda y}{(\lambda x)^2 + (\lambda y)^2} = \frac{\lambda(x - y)}{\lambda^2(x^2 + y^2)} = \lambda^{-1} f(x, y)$. This is homogeneous of degree $-1$.
$(B)$ $f(\lambda x, \lambda y) = (\lambda x)^{1/3} (\lambda y)^{-2/3} \tan^{-1} \frac{\lambda x}{\lambda y} = \lambda^{1/3 - 2/3} x^{1/3} y^{-2/3} \tan^{-1} \frac{x}{y} = \lambda^{-1/3} f(x, y)$. This is homogeneous of degree $-1/3$.
$(C)$ $f(\lambda x, \lambda y) = \lambda x (\ln \sqrt{\lambda^2(x^2 + y^2)} - \ln(\lambda y)) + \lambda y e^{\frac{\lambda x}{\lambda y}} = \lambda x (\ln \lambda + \ln \sqrt{x^2 + y^2} - \ln \lambda - \ln y) + \lambda y e^{x/y} = \lambda f(x, y)$. This is homogeneous of degree $1$.
$(D)$ $f(\lambda x, \lambda y) = \lambda x [\ln \frac{2\lambda^2 x^2 + \lambda^2 y^2}{\lambda x} - \ln(\lambda x + \lambda y)] + (\lambda y)^2 \tan \frac{\lambda x + 2\lambda y}{3\lambda x - \lambda y} = \lambda x [\ln \frac{\lambda^2(2x^2 + y^2)}{\lambda x} - \ln(\lambda(x + y))] + \lambda^2 y^2 \tan \frac{x + 2y}{3x - y} = \lambda x [\ln \lambda + \ln \frac{2x^2 + y^2}{x(x + y)}] + \lambda^2 y^2 \tan \frac{x + 2y}{3x - y}$. Since the $\ln \lambda$ term does not factor out as $\lambda^n$,this function is not homogeneous.

(A) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} - \cos^2 \left( \frac{y}{x} \right)$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = v - \cos^2 v$.
Simplifying,we get: $x \frac{dv}{dx} = -\cos^2 v$.
Separating the variables: $\frac{dv}{\cos^2 v} = -\frac{dx}{x}$,which is $\sec^2 v \, dv = -\frac{dx}{x}$.
Integrating both sides: $\int \sec^2 v \, dv = -\int \frac{1}{x} \, dx$.
This gives: $\tan v = -\ln |x| + C$.
Substituting $v = \frac{y}{x}$: $\tan \left( \frac{y}{x} \right) = -\ln x + C$.
The curve passes through $(1, \frac{\pi}{4})$,so $\tan \left( \frac{\pi/4}{1} \right) = -\ln(1) + C$.
Since $\tan(\frac{\pi}{4}) = 1$ and $\ln(1) = 0$,we have $1 = 0 + C$,so $C = 1$.
Thus,$\tan \left( \frac{y}{x} \right) = 1 - \ln x = \ln e - \ln x = \ln \left( \frac{e}{x} \right)$.
Therefore,$y = x \tan^{-1} \left( \ln \frac{e}{x} \right)$.

(C) Given the differential equation: $2x^4y \frac{dy}{dx} + y^4 = 4x^6$.
Substitute $y = u^m$,then $\frac{dy}{dx} = m u^{m-1} \frac{du}{dx}$.
Substituting these into the equation: $2x^4(u^m)(m u^{m-1} \frac{du}{dx}) + (u^m)^4 = 4x^6$.
This simplifies to: $2m x^4 u^{2m-1} \frac{du}{dx} + u^{4m} = 4x^6$.
For the equation to be homogeneous,the powers of $x$ and $u$ must be proportional in each term.
Comparing the powers of $x$ and $u$ in the terms $x^4 u^{2m-1} \frac{du}{dx}$,$u^{4m}$,and $x^6$:
From $u^{4m}$ and $x^6$,we require $4m = 6$,which gives $m = 3/2$.
Checking the first term $x^4 u^{2m-1}$,we require $2m-1 = 2$,which also gives $m = 3/2$.
Thus,the value of $m$ is $3/2$.

(A) Let the point $P$ be $(x, y)$. The slope of the tangent at $P$ is $m = \frac{dy}{dx}$. The slope of the normal at $P$ is $-\frac{1}{m} = -\frac{dx}{dy}$.
The equation of the normal at $P(x, y)$ is $Y - y = -\frac{dx}{dy} (X - x)$.
To find the $x$-intercept,set $Y = 0$:
$-y = -\frac{dx}{dy} (X - x) \implies X - x = y \frac{dy}{dx} \implies X = x + y \frac{dy}{dx}$.
The abscissa of $P$ is $x$. The $x$-intercept of the normal is $X = x + y \frac{dy}{dx}$.
The radius vector of $P$ is $\sqrt{x^2 + y^2}$,so its square is $x^2 + y^2$.
According to the problem,the product of the abscissa and the $x$-intercept equals twice the square of the radius vector:
$x \left( x + y \frac{dy}{dx} \right) = 2(x^2 + y^2)$.
$x^2 + xy \frac{dy}{dx} = 2x^2 + 2y^2 \implies xy \frac{dy}{dx} = x^2 + 2y^2$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
$x(vx) (v + x \frac{dv}{dx}) = x^2 + 2(vx)^2 \implies x^2 v (v + x \frac{dv}{dx}) = x^2(1 + 2v^2)$.
$v^2 + vx \frac{dv}{dx} = 1 + 2v^2 \implies vx \frac{dv}{dx} = 1 + v^2$.
$\frac{v}{1 + v^2} dv = \frac{1}{x} dx$.
Integrating both sides: $\frac{1}{2} \ln(1 + v^2) = \ln x + C \implies \ln(1 + v^2) = \ln x^2 + 2C \implies 1 + v^2 = k x^2$,where $k = e^{2C}$.
Substituting $v = \frac{y}{x}$: $1 + \frac{y^2}{x^2} = k x^2 \implies x^2 + y^2 = k x^4$.
The curve passes through $(1, 0)$,so $1^2 + 0^2 = k(1)^4 \implies k = 1$.
Thus,the equation is $x^2 + y^2 = x^4$.

(A) Given the differential equation $(x^2y^2 - 1)dy + 2xy^3dx = 0$.
Substitute $y = z^{\alpha}$,then $dy = \alpha z^{\alpha - 1} dz$.
Substituting these into the equation:
$(x^2(z^{\alpha})^2 - 1)(\alpha z^{\alpha - 1} dz) + 2x(z^{\alpha})^3 dx = 0$
$(x^2 z^{2\alpha} - 1)(\alpha z^{\alpha - 1} dz) + 2x z^{3\alpha} dx = 0$
$\alpha x^2 z^{3\alpha - 1} dz - \alpha z^{\alpha - 1} dz + 2x z^{3\alpha} dx = 0$
For the equation to be homogeneous,the sum of powers of $x$ and $z$ in each term must be equal.
The powers are:
Term $1$: $x^2 z^{3\alpha - 1} \rightarrow 2 + (3\alpha - 1) = 3\alpha + 1$
Term $2$: $z^{\alpha - 1} \rightarrow 0 + (\alpha - 1) = \alpha - 1$
Term $3$: $x z^{3\alpha} \rightarrow 1 + 3\alpha = 3\alpha + 1$
Equating the powers: $3\alpha + 1 = \alpha - 1$
$2\alpha = -2 \Rightarrow \alpha = -1$.

(B) Given the differential equation: $x \frac{dy}{dx} = y \ln \left( \frac{y}{x} \right)$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $x(v + x \frac{dv}{dx}) = vx \ln(v)$.
Dividing by $x$: $v + x \frac{dv}{dx} = v \ln(v)$.
Rearranging: $x \frac{dv}{dx} = v \ln(v) - v = v(\ln(v) - 1)$.
Separating variables: $\frac{dv}{v(\ln(v) - 1)} = \frac{dx}{x}$.
Integrating both sides: $\int \frac{dv}{v(\ln(v) - 1)} = \int \frac{dx}{x}$.
Let $u = \ln(v) - 1$,then $du = \frac{1}{v} dv$.
So,$\int \frac{du}{u} = \ln|x| + C$.
$\ln|\ln(v) - 1| = \ln|x| + C$.
$\ln(v) - 1 = Cx$ (where $C$ is a constant).
$\ln(\frac{y}{x}) = 1 + Cx$.
$\frac{y}{x} = e^{1 + Cx}$.
$y = xe^{1 + Cx}$.

(D) Step $1$: Check option $A$. $A$ function $f(x, y)$ is homogeneous of degree $n$ if $f(\lambda x, \lambda y) = \lambda^n f(x, y)$. For $f(x, y) = e^{y/x} + \tan(y/x)$,$f(\lambda x, \lambda y) = e^{(\lambda y)/(\lambda x)} + \tan((\lambda y)/(\lambda x)) = e^{y/x} + \tan(y/x) = \lambda^0 f(x, y)$. Thus,it is homogeneous of degree $0$. Option $A$ is true.
Step $2$: Check option $B$. $A$ differential equation $M(x, y)dx + N(x, y)dy = 0$ is homogeneous if both $M(x, y)$ and $N(x, y)$ are homogeneous functions of the same degree. Here,$M(x, y) = x \ln(y/x)$ is homogeneous of degree $1$ (since $M(\lambda x, \lambda y) = \lambda x \ln(y/x) = \lambda^1 M(x, y)$). $N(x, y) = \frac{y^2}{x} \sin^{-1}(y/x)$ is homogeneous of degree $1$ (since $N(\lambda x, \lambda y) = \frac{(\lambda y)^2}{\lambda x} \sin^{-1}(y/x) = \lambda \frac{y^2}{x} \sin^{-1}(y/x) = \lambda^1 N(x, y)$). Since both are degree $1$,the equation is homogeneous. Option $B$ is true.
Step $3$: Check option $C$. For $f(x, y) = x^2 + \sin x \cos y$,$f(\lambda x, \lambda y) = (\lambda x)^2 + \sin(\lambda x) \cos(\lambda y) = \lambda^2 x^2 + \sin(\lambda x) \cos(\lambda y)$. This cannot be written in the form $\lambda^n f(x, y)$ for any $n$. Thus,it is not homogeneous. Option $C$ is true.
Step $4$: Since $A, B,$ and $C$ are true,the correct answer is $D$.

(D) function $f(x, y)$ is homogeneous of degree $n$ if $f(\lambda x, \lambda y) = \lambda^n f(x, y)$.
$A) f(x, y) = x \sin y + y \sin x$
$f(\lambda x, \lambda y) = \lambda x \sin(\lambda y) + \lambda y \sin(\lambda x) = \lambda(x \sin(\lambda y) + y \sin(\lambda x))$.
Since this is not equal to $\lambda^n f(x, y)$,it is not homogeneous.
$B) f(x, y) = x e^{y/x} + y e^{x/y}$
$f(\lambda x, \lambda y) = \lambda x e^{\lambda y / \lambda x} + \lambda y e^{\lambda x / \lambda y} = \lambda(x e^{y/x} + y e^{x/y}) = \lambda^1 f(x, y)$.
This is a homogeneous function of degree $1$.
$C) f(x, y) = x^2 - xy$
$f(\lambda x, \lambda y) = (\lambda x)^2 - (\lambda x)(\lambda y) = \lambda^2 x^2 - \lambda^2 xy = \lambda^2(x^2 - xy) = \lambda^2 f(x, y)$.
This is a homogeneous function of degree $2$.
Therefore,both $(B)$ and $(C)$ are homogeneous functions.

(A) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} - \cos^2 \left( \frac{y}{x} \right)$.
Let $v = \frac{y}{x}$,then $y = vx$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting into the equation: $v + x \frac{dv}{dx} = v - \cos^2 v$.
This simplifies to $x \frac{dv}{dx} = -\cos^2 v$.
Separating variables: $\sec^2 v \, dv = -\frac{dx}{x}$.
Integrating both sides: $\int \sec^2 v \, dv = -\int \frac{dx}{x} \implies \tan v = -\ln |x| + C$.
Since the curve passes through $\left( 1, \frac{\pi}{4} \right)$,we have $v = \frac{\pi}{4}$ when $x = 1$.
$\tan \left( \frac{\pi}{4} \right) = -\ln(1) + C \implies 1 = 0 + C \implies C = 1$.
Thus,$\tan v = 1 - \ln x = \ln e - \ln x = \ln \left( \frac{e}{x} \right)$.
Therefore,$v = \tan^{-1} \left( \ln \frac{e}{x} \right)$.
Substituting $v = \frac{y}{x}$,we get $y = x \tan^{-1} \left( \ln \frac{e}{x} \right)$.

(B) The given differential equation is $(x^2 - xy)dy = (xy + y^2)dx$.
This can be written as $\frac{dy}{dx} = \frac{xy + y^2}{x^2 - xy}$.
This is a homogeneous differential equation.
Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{x(vx) + (vx)^2}{x^2 - x(vx)} = \frac{v + v^2}{1 - v}$.
$x\frac{dv}{dx} = \frac{v + v^2}{1 - v} - v = \frac{v + v^2 - v + v^2}{1 - v} = \frac{2v^2}{1 - v}$.
Separating the variables: $\frac{1 - v}{v^2} dv = \frac{2}{x} dx$.
Integrating both sides: $\int (v^{-2} - v^{-1}) dv = 2 \int \frac{1}{x} dx$.
$-v^{-1} - \ln|v| = 2\ln|x| + C$.
Substituting $v = y/x$: $-\frac{x}{y} - \ln|y/x| = 2\ln|x| + C$.
$-\frac{x}{y} - \ln|y| + \ln|x| = 2\ln|x| + C$.
$-\frac{x}{y} = \ln|y| + \ln|x| + C = \ln|xy| + C$.
Thus,$xy = e^{-x/y - C} = Ce^{-x/y}$.

(B) Given equation: $x \cos \left( \frac{y}{x} \right) (y dx + x dy) = y \sin \left( \frac{y}{x} \right) (x dy - y dx)$
Divide both sides by $xy \cos \left( \frac{y}{x} \right)$:
$\frac{y dx + x dy}{y} = \tan \left( \frac{y}{x} \right) \frac{x dy - y dx}{x} $
$\frac{y dx + x dy}{xy} = \tan \left( \frac{y}{x} \right) \left( \frac{x dy - y dx}{x^2} \right) $
Note that $d(xy) = y dx + x dy$ and $d\left( \frac{y}{x} \right) = \frac{x dy - y dx}{x^2}$.
So,$\frac{d(xy)}{xy} = \tan \left( \frac{y}{x} \right) d\left( \frac{y}{x} \right) $
Integrating both sides:
$\int \frac{d(xy)}{xy} = \int \tan \left( \frac{y}{x} \right) d\left( \frac{y}{x} \right) $
$\ln |xy| = \ln \left| \sec \left( \frac{y}{x} \right) \right| + \ln |c| $
$xy = c \sec \left( \frac{y}{x} \right) $
Thus,$x = \frac{c}{y} \sec \left( \frac{y}{x} \right)$ is not the form,but $xy = c \sec \left( \frac{y}{x} \right)$ is equivalent to $xy \cos \left( \frac{y}{x} \right) = c$.

(D) Given the differential equation: $x dy - y dx = \sqrt{x^2 - y^2} dx$.
Divide by $dx$: $x \frac{dy}{dx} - y = \sqrt{x^2 - y^2}$.
Substitute $y = vx$,so $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
$x(v + x \frac{dv}{dx}) - vx = \sqrt{x^2 - v^2x^2}$.
$vx + x^2 \frac{dv}{dx} - vx = x \sqrt{1 - v^2}$.
$x^2 \frac{dv}{dx} = x \sqrt{1 - v^2} \implies \frac{dv}{\sqrt{1 - v^2}} = \frac{dx}{x}$.
Integrating both sides: $\int \frac{dv}{\sqrt{1 - v^2}} = \int \frac{dx}{x} \implies \sin^{-1}(v) = \ln|x| + C$.
Substitute $v = \frac{y}{x}$: $\sin^{-1}(\frac{y}{x}) = \ln|x| + C$.
Given $y(1) = 0$,at $x = 1, y = 0$: $\sin^{-1}(0) = \ln(1) + C \implies 0 = 0 + C \implies C = 0$.
Thus,$\sin^{-1}(\frac{y}{x}) = \ln x \implies \frac{y}{x} = \sin(\ln x) \implies y = x \sin(\ln x)$.

(B) Given differential equation is $(x^2 - y^2) \, dx + 2xy \, dy = 0$.
This can be written as $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$.
Since this is a homogeneous differential equation,we put $y = ux$,which implies $\frac{dy}{dx} = u + x \frac{du}{dx}$.
Substituting these into the equation: $u + x \frac{du}{dx} = \frac{u^2 x^2 - x^2}{2x(ux)} = \frac{u^2 - 1}{2u}$.
$x \frac{du}{dx} = \frac{u^2 - 1}{2u} - u = \frac{u^2 - 1 - 2u^2}{2u} = \frac{-(1 + u^2)}{2u}$.
Separating variables: $\int \frac{2u}{1 + u^2} \, du = - \int \frac{1}{x} \, dx$.
Integrating both sides: $\ln(1 + u^2) = -\ln|x| + \ln|C|$.
$\ln(1 + u^2) = \ln\left(\frac{C}{x}\right) \Rightarrow 1 + u^2 = \frac{C}{x}$.
Substituting $u = \frac{y}{x}$: $1 + \frac{y^2}{x^2} = \frac{C}{x} \Rightarrow \frac{x^2 + y^2}{x^2} = \frac{C}{x} \Rightarrow x^2 + y^2 = Cx$.
Since the curve passes through $(1, 1)$,we have $1^2 + 1^2 = C(1) \Rightarrow C = 2$.
Thus,the equation of the curve is $x^2 + y^2 = 2x$,which can be rewritten as $(x - 1)^2 + y^2 = 1$.
This represents a circle with center $(1, 0)$ and radius $1$.

(B) Given differential equation is $\frac{dy}{dx} = \frac{y^3}{2(xy^2 - x^2)}$.
Let $z = y^2$. Then $\frac{dz}{dx} = 2y \frac{dy}{dx}$.
Substituting $\frac{dy}{dx} = \frac{1}{2y} \frac{dz}{dx}$ into the equation:
$\frac{1}{2y} \frac{dz}{dx} = \frac{y^3}{2(xy^2 - x^2)} \implies \frac{dz}{dx} = \frac{y^4}{xy^2 - x^2} = \frac{z^2}{xz - x^2}$.
Dividing numerator and denominator by $x^2$,we get $\frac{dz}{dx} = \frac{(z/x)^2}{(z/x) - 1}$. This is a homogeneous differential equation in terms of $z$ and $x$. Thus,Statement $-1$ is true.
To solve $\frac{dz}{dx} = \frac{z^2}{xz - x^2}$,let $z = vx$,then $\frac{dz}{dx} = v + x \frac{dv}{dx}$.
$v + x \frac{dv}{dx} = \frac{v^2 x^2}{x(vx) - x^2} = \frac{v^2}{v - 1}$.
$x \frac{dv}{dx} = \frac{v^2}{v - 1} - v = \frac{v^2 - v^2 + v}{v - 1} = \frac{v}{v - 1}$.
$\int \frac{v - 1}{v} dv = \int \frac{1}{x} dx \implies \int (1 - \frac{1}{v}) dv = \ln|x| + C$.
$v - \ln|v| = \ln|x| + C \implies \frac{z}{x} - \ln|\frac{z}{x}| = \ln|x| + C$.
$\frac{y^2}{x} - \ln|y^2| + \ln|x| = \ln|x| + C \implies \frac{y^2}{x} = \ln|y^2| + C$.
This does not match $y^2 e^{-y^2/x} = C$. Thus,Statement $-2$ is false.

(D) Given the differential equation $\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$.
Substitute $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting into the equation: $v + x \frac{dv}{dx} = \frac{x(vx)}{x^2 + v^2 x^2} = \frac{v}{1 + v^2}$.
$x \frac{dv}{dx} = \frac{v}{1 + v^2} - v = \frac{v - v - v^3}{1 + v^2} = -\frac{v^3}{1 + v^2}$.
Separating variables: $\int \frac{1 + v^2}{v^3} dv = -\int \frac{dx}{x}$.
$\int (v^{-3} + v^{-1}) dv = -\int \frac{dx}{x}$.
Integrating both sides: $-\frac{1}{2v^2} + \ln|v| = -\ln|x| + C$.
Substitute $v = \frac{y}{x}$: $-\frac{x^2}{2y^2} + \ln|\frac{y}{x}| = -\ln|x| + C$.
$-\frac{x^2}{2y^2} + \ln|y| - \ln|x| = -\ln|x| + C \implies -\frac{x^2}{2y^2} + \ln|y| = C$.
Using $y(1) = 1$: $-\frac{1^2}{2(1)^2} + \ln(1) = C \implies C = -\frac{1}{2}$.
The equation is $-\frac{x^2}{2y^2} + \ln|y| = -\frac{1}{2}$.
For $y = e$: $-\frac{x^2}{2e^2} + \ln(e) = -\frac{1}{2} \implies -\frac{x^2}{2e^2} + 1 = -\frac{1}{2}$.
$-\frac{x^2}{2e^2} = -\frac{3}{2} \implies x^2 = 3e^2 \implies x = \sqrt{3}e$.

(NONE) In a deck of $52$ cards,there are $4$ kings,$4$ queens,and $4$ jacks.
$P(E) = P(\text{king or queen}) = \frac{4+4}{52} = \frac{8}{52} = \frac{2}{13}$.
$P(F) = P(\text{queen or jack}) = \frac{4+4}{52} = \frac{8}{52} = \frac{2}{13}$.
The event $E \cap F$ represents the card being a queen (since queen is common to both sets).
$P(E \cap F) = P(\text{queen}) = \frac{4}{52} = \frac{1}{13}$.
Now,check for independence: $P(E) \times P(F) = \frac{2}{13} \times \frac{2}{13} = \frac{4}{169}$.
Since $P(E \cap F) = \frac{1}{13} = \frac{13}{169}$,and $\frac{4}{169} \neq \frac{13}{169}$,we have $P(E \cap F) \neq P(E) \times P(F)$.
Therefore,the events $E$ and $F$ are not independent.

(N/A) The given differential equation can be expressed as
$\frac{dy}{dx} = \frac{x+2y}{x-y}$ ............$(1)$
Let $F(x, y) = \frac{x+2y}{x-y}$.
Now $F(\lambda x, \lambda y) = \frac{\lambda(x+2y)}{\lambda(x-y)} = \lambda^0 \cdot F(x, y)$.
Therefore,$F(x, y)$ is a homogeneous function of degree zero. So,the given differential equation is a homogeneous differential equation.
Alternatively,
$\frac{dy}{dx} = \frac{1+2(y/x)}{1-(y/x)} = g(y/x)$ .............$(2)$
Since the $R.H.S.$ is a function of $y/x$,it is a homogeneous function of degree zero.
To solve it,we make the substitution $y = vx$ ...........$(3)$
Differentiating with respect to $x$,we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$ ...........$(4)$
Substituting $(3)$ and $(4)$ into $(1)$:
$v + x \frac{dv}{dx} = \frac{1+2v}{1-v}$
$x \frac{dv}{dx} = \frac{1+2v}{1-v} - v = \frac{1+2v-v+v^2}{1-v} = \frac{v^2+v+1}{1-v}$
$\frac{v-1}{v^2+v+1} dv = -\frac{dx}{x}$
Integrating both sides:
$\int \frac{v-1}{v^2+v+1} dv = -\int \frac{dx}{x}$
$\frac{1}{2} \int \frac{2v+1-3}{v^2+v+1} dv = -\log|x| + C_1$
$\frac{1}{2} \log|v^2+v+1| - \frac{3}{2} \int \frac{1}{(v+1/2)^2 + (\sqrt{3}/2)^2} dv = -\log|x| + C_1$
$\frac{1}{2} \log|v^2+v+1| - \frac{3}{2} \cdot \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2v+1}{\sqrt{3}}\right) = -\log|x| + C_1$
$\frac{1}{2} \log|v^2+v+1| + \log|x| = \sqrt{3} \tan^{-1}\left(\frac{2v+1}{\sqrt{3}}\right) + C_1$
$\frac{1}{2} \log|x^2(v^2+v+1)| = \sqrt{3} \tan^{-1}\left(\frac{2v+1}{\sqrt{3}}\right) + C_1$
Substituting $v = y/x$:
$\log|x^2+xy+y^2| = 2\sqrt{3} \tan^{-1}\left(\frac{2y+x}{\sqrt{3}x}\right) + C$.

(N/A) The given differential equation can be written as
$\frac{d y}{d x}=\frac{y \cos \left(\frac{y}{x}\right)+x}{x \cos \left(\frac{y}{x}\right)}$ ............$(1)$
It is a differential equation of the form $\frac{d y}{d x}=F(x, y)$.
Here $F(x, y) = \frac{y \cos \left(\frac{y}{x}\right) + x}{x \cos \left(\frac{y}{x}\right)}$.
Replacing $x$ by $\lambda x$ and $y$ by $\lambda y$,we get
$F(\lambda x, \lambda y) = \frac{\lambda y \cos \left(\frac{\lambda y}{\lambda x}\right) + \lambda x}{\lambda x \cos \left(\frac{\lambda y}{\lambda x}\right)} = \frac{\lambda [y \cos (y/x) + x]}{\lambda [x \cos (y/x)]} = \lambda^0 F(x, y)$.
Since $F(x, y)$ is a homogeneous function of degree zero,the given differential equation is homogeneous.
To solve it,substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into equation $(1)$:
$v + x \frac{dv}{dx} = \frac{vx \cos v + x}{x \cos v} = \frac{v \cos v + 1}{\cos v}$.
$x \frac{dv}{dx} = \frac{v \cos v + 1}{\cos v} - v = \frac{v \cos v + 1 - v \cos v}{\cos v} = \frac{1}{\cos v}$.
Separating variables: $\cos v \, dv = \frac{dx}{x}$.
Integrating both sides: $\int \cos v \, dv = \int \frac{1}{x} \, dx$.
$\sin v = \log |x| + C$.
Substituting $v = y/x$,the general solution is $\sin \left(\frac{y}{x}\right) = \log |x| + C$.