(a) Homogeneous ODE $\to y=vx\to$ solution $xy(x-y)=C$. (b) Fisher index $\approx110.71$; TRT and FRT both satisfied.

Part (a) — solve $(y^{2}-2xy)\,dx=(x^{2}-2xy)\,dy$. This is homogeneous:

$\frac{dy}{dx}=\frac{y^{2}-2xy}{x^{2}-2xy}.$

Put $y=vx\Rightarrow \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$:

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

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

Separate variables:

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

Partial fractions: $\dfrac{1-2v}{v(v-1)}=\dfrac{-1}{v}+\dfrac{-1}{v-1}$, so

$\frac13\int\!\left(-\frac1v-\frac1{v-1}\right)dv=\int\frac{dx}{x}\ \Rightarrow\ -\frac13\ln|v(v-1)|=\ln|x|+\text{const}.$

Hence $v(v-1)=\dfrac{C}{x^{3}}$. With $v=\dfrac{y}{x}$, $v(v-1)=\dfrac{y(y-x)}{x^{2}}$, giving

$\frac{y(y-x)}{x^{2}}=\frac{C}{x^{3}}\ \Rightarrow\ x\,y(y-x)=C\ \Rightarrow\ x^{2}y-xy^{2}=C.$

Check: $d(x^{2}y-xy^{2})=(2xy-y^{2})dx+(x^{2}-2xy)dy=0$ reproduces $(y^{2}-2xy)dx=(x^{2}-2xy)dy$. ✓

Part (b) — Fisher's price index and reversal tests. Compute the four aggregates.

$\sum p_0q_0=1220,\ \sum p_1q_0=1341,\ \sum p_0q_1=1600,\ \sum p_1q_1=1784.$

Fisher's price index:

$P_{01}^{F}=\sqrt{\frac{\sum p_1q_0}{\sum p_0q_0}\times\frac{\sum p_1q_1}{\sum p_0q_1}}\times100=\sqrt{\frac{1341}{1220}\times\frac{1784}{1600}}\times100.$

$=\sqrt{1.0992\times1.115}\times100=\sqrt{1.2256}\times100=1.1071\times100=110.71.$

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