Q.(a) Solve the following differential equation.
$\dfrac{dy}{dx}=\dfrac{x^{2}+y^{2}}{xy}$

(a) Substituting $y=vx$ separates variables and gives $y^2=x^2(2\log|x|+C)$. (b) With $n=4,p=1/2$: $703$, $516$ and $656$ families respectively.

Part (a) — Homogeneous DE $\dfrac{dy}{dx}=\dfrac{x^{2}+y^{2}}{xy}.$

Put $y=vx\Rightarrow \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}.$ The right side becomes $\dfrac{x^{2}+v^{2}x^{2}}{x\cdot vx}=\dfrac{1+v^{2}}{v}.$

$v+x\frac{dv}{dx}=\frac{1+v^{2}}{v}\Rightarrow x\frac{dv}{dx}=\frac{1+v^{2}}{v}-v=\frac{1}{v}.$

Separate variables: $v\,dv=\dfrac{dx}{x}.$ Integrate:

$\frac{v^{2}}{2}=\log|x|+c\Rightarrow v^{2}=2\log|x|+C.$

Replace $v=\dfrac{y}{x}$:

$\frac{y^{2}}{x^{2}}=2\log|x|+C\Rightarrow y^{2}=x^{2}\big(2\log|x|+C\big).$

Part (b) — Binomial, boys per family $X\sim B(4,\tfrac12)$, $P(X=r)=\binom{4}{r}\left(\tfrac12\right)^{4}=\dfrac{\binom{4}{r}}{16}.$ Total families $=750.$

(i) At least one boy $=1-P(0)=1-\dfrac{1}{16}=\dfrac{15}{16}.$ Families $=750\times\dfrac{15}{16}=703.125\approx703.$

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