Homogeneous differential equations Questions in English

(A) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} + \tan \frac{y}{x}$.
This is a homogeneous differential equation.
Let $v = \frac{y}{x}$,so $y = vx$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = v + \tan v$.
Subtracting $v$ from both sides: $x \frac{dv}{dx} = \tan v$.
Separating the variables: $\frac{dv}{\tan v} = \frac{dx}{x}$,which is $\cot v \, dv = \frac{dx}{x}$.
Integrating both sides: $\int \cot v \, dv = \int \frac{dx}{x}$.
This gives $\ln |\sin v| = \ln |x| + \ln |c|$.
Using logarithmic properties: $\ln |\sin v| = \ln |cx|$.
Taking the exponential of both sides: $\sin v = cx$.
Substituting back $v = \frac{y}{x}$: $\sin(\frac{y}{x}) = cx$,or $x = c \sin(\frac{y}{x})$.

(C) Given the differential equation: $y \frac{dy}{dx} = x \left[ \frac{y^2}{x^2} + \frac{\phi(y^2/x^2)}{\phi'(y^2/x^2)} \right]$.
Dividing by $y$,we get $\frac{dy}{dx} = \frac{y}{x} + \frac{x \phi(y^2/x^2)}{y \phi'(y^2/x^2)}$.
Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = v + \frac{x \phi(v^2)}{vx \phi'(v^2)} = v + \frac{\phi(v^2)}{v \phi'(v^2)}$.
Subtracting $v$ from both sides: $x \frac{dv}{dx} = \frac{\phi(v^2)}{v \phi'(v^2)}$.
Separating variables: $\frac{v \phi'(v^2)}{\phi(v^2)} dv = \frac{dx}{x}$.
Integrating both sides: $\int \frac{v \phi'(v^2)}{\phi(v^2)} dv = \int \frac{dx}{x}$.
Let $u = v^2$,then $du = 2v dv$,so $v dv = \frac{1}{2} du$.
The integral becomes $\frac{1}{2} \int \frac{\phi'(u)}{\phi(u)} du = \ln|x| + C_1$.
$\frac{1}{2} \ln|\phi(u)| = \ln|x| + C_1 \Rightarrow \ln|\phi(v^2)| = 2 \ln|x| + 2C_1 = \ln|x^2| + \ln|c|$.
Thus,$\phi(v^2) = c x^2$.
Substituting $v^2 = y^2/x^2$,we get $\phi(y^2/x^2) = c x^2$.

(C) Let $B$ denote a boy and $G$ denote a girl. The possible outcomes for $2$ children are $\{BB, BG, GB, GG\}$,where each outcome is equally likely with probability $1/4$.
Given that at least $1$ child is a boy,the sample space reduces to $S = \{BB, BG, GB\}$.
The total number of outcomes in the reduced sample space is $n(S) = 3$.
We want to find the probability that the other child is a girl,which corresponds to the outcomes where there is exactly $1$ boy and $1$ girl. These outcomes are $\{BG, GB\}$.
The number of favourable outcomes is $n(E) = 2$.
Therefore,the required probability is $P(E) = \frac{n(E)}{n(S)} = \frac{2}{3}$.

(A) The given differential equation is $x \sin(\frac{y}{x}) dy = (y \sin(\frac{y}{x}) - x) dx$.
Substituting $y = vx$,we get $dy = v dx + x dv$.
Substituting into the equation: $x \sin v (v dx + x dv) = (vx \sin v - x) dx$.
$vx \sin v dx + x^2 \sin v dv = vx \sin v dx - x dx$.
$x^2 \sin v dv = -x dx \Rightarrow \sin v dv = -\frac{1}{x} dx$.
Integrating both sides: $-\cos v = -\ln|x| + C$.
Given $y(1) = \frac{\pi}{2}$,we have $v(1) = \frac{\pi}{2}$.
$-\cos(\frac{\pi}{2}) = -\ln(1) + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$.
Thus,$\cos(\frac{y}{x}) = \ln x$.
Given $\alpha = \cos(\frac{e^{12}}{e^{12}}) = \cos(1)$.
The circle equation is $x^2 + y^2 - 2px + 2py + \alpha + 2 = 0$.
The radius $r$ is given by $\sqrt{p^2 + (-p)^2 - (\alpha + 2)} = \sqrt{2p^2 - \alpha - 2}$.
We require $r \leq 6$,so $2p^2 - \alpha - 2 \leq 36$.
$2p^2 \leq 38 + \alpha$.
Since $\alpha = \cos(1) \approx 0.54$,$2p^2 \leq 38.54 \Rightarrow p^2 \leq 19.27$.
The possible integral values for $p$ are $p \in \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$.
There are $9$ such values.