(a) Homogeneous DE, $y=vx$ $\Rightarrow$ $x^2y^2(y^2-x^2)=12$. (b) Quarterly averages $243.83,312,361.67,386.83$; grand mean $326.08$; indices $74.78,95.68,110.91,118.63$.

(a) Equation of the curve.

The equation $(y^3-2yx^2)\,dx+(2xy^2-x^3)\,dy=0$ is homogeneous (each term degree $3$). Put $y=vx$, $dy=v\,dx+x\,dv$:

$x^3(v^3-2v)\,dx+x^3(2v^2-1)(v\,dx+x\,dv)=0.$

Divide by $x^3$ and collect $dx$ terms:

$\big[(v^3-2v)+(2v^3-v)\big]\,dx+(2v^2-1)x\,dv=0\Rightarrow 3v(v^2-1)\,dx+(2v^2-1)x\,dv=0.$

Separate:

$3\,\frac{dx}{x}=-\frac{2v^2-1}{v(v^2-1)}\,dv.$

By partial fractions $\dfrac{2v^2-1}{v(v-1)(v+1)}=\dfrac{1}{v}+\dfrac{1/2}{v-1}+\dfrac{1/2}{v+1}$, so

$3\ln x=-\Big[\ln v+\tfrac12\ln(v^2-1)\Big]+c.$

Multiply by $2$: $6\ln x+2\ln v+\ln(v^2-1)=c'$, i.e. $x^6v^2(v^2-1)=C.$

With $v=y/x$: $x^6\cdot\dfrac{y^2}{x^2}\cdot\dfrac{y^2-x^2}{x^2}=x^2y^2(y^2-x^2)=C.$

At $(1,2)$: $1\cdot4\cdot(4-1)=12$, so $C=12.$ Hence the curve is

$x^2y^2(y^2-x^2)=12.$

(b) Seasonal index — method of simple averages.

Average each quarter over the six years: