We have:
\begin{align*}\frac {\mathrm dz} {\mathrm dx} &= ny^{n - 1} \frac {\mathrm dy} {\mathrm dx} \times \frac {\mathrm dy} {\mathrm dx} + 2 y^n \frac {\mathrm d^2 y} {\mathrm dx} \frac {\mathrm dy} {\mathrm dx} \\ & = y^{n - 1} \frac {\mathrm dy} {\mathrm dx} \left(n \left(\frac {\mathrm dy} {\mathrm dx}\right)^2 + 2y \frac {\mathrm d^2 y} {\mathrm dx^2}\right)\end{align*}
(i) Note that setting n = 1 we have:
\displaystyle \frac {\frac {\mathrm dz} {\mathrm dx}} {\frac {\mathrm dy} {\mathrm dx}} = \frac {\mathrm dz} {\mathrm dy} = \sqrt y
where \displaystyle z = y \left(\frac {\mathrm dy} {\mathrm dx}\right). We have:
\displaystyle z = \frac 2 3 y^{3/2} + C
ie.
\displaystyle y \left(\frac {\mathrm dy} {\mathrm dx}\right)^2 = \frac 2 3 y^{3/2} +C
when x = 0 we have 0 = \frac 2 3 + C \implies C = -\frac 2 3. So:
\displaystyle \frac {\sqrt y} {\sqrt {y^{3/2} - 1}} \frac {\mathrm dy} {\mathrm dx} = \sqrt {\frac 2 3}
We have:
\displaystyle \int \frac {\sqrt y} {\sqrt {y^{3/2} - 1}} \mathrm dy = \frac 4 3 (y^{3/2} - 1)^{1/2} + C
so:
\displaystyle \frac 4 3 (y^{3/2} - 1)^{1/2} = \sqrt {\frac 2 3} x + C
At x = 0 we have y = 1 so C = 0. So:
\displaystyle \frac 4 3 (y^{3/2} - 1)^{1/2} = \sqrt {\frac 2 3} x
Squaring:
\displaystyle \frac {16} 9 (y^{3/2} - 1) = \frac 2 3 x^2
So:
\displaystyle (y^{3/2} - 1) = \left(\frac 2 3 \times \frac 9 {16}\right) x^2 = \frac 3 8 x^2.
Finally giving:
\displaystyle y = \left(\frac 3 8 x^2 + 1\right)^{2/3}
(ii)
We want to get the LHS into something involving \displaystyle \frac {\mathrm dy} {\mathrm dx}, note that we can rewrite the equation as:
\displaystyle \left(\frac {\mathrm dy} {\mathrm dx}\right)^2 - y \frac {\mathrm d^2 y} {\mathrm dx^2} = -y^2
Multiplying by -2 we have:
\displaystyle -2 \left(\frac {\mathrm dy} {\mathrm dx}\right)^2 + 2y \frac {\mathrm d^2 y} {\mathrm dx^2} = 2y^2
This suggests we want to use n = -2, that is: \displaystyle z = y^{-2} \left(\frac {\mathrm dy} {\mathrm dx}\right)^2, which has:
\displaystyle \frac {\mathrm dz} {\mathrm dx} = \frac 1 {y^3} \frac {\mathrm dy} {\mathrm dx} \left(-2 \left(\frac {\mathrm dy} {\mathrm dx}\right)^2 + 2y \frac {\mathrm d^2 y} {\mathrm dx^2}\right)
So the LHS of the original equation is equal to:
\displaystyle y^3 \frac {\mathrm dz} {\mathrm dy}
giving:
\displaystyle y^3 \frac {\mathrm dz} {\mathrm dy} = 2y^2
So:
\displaystyle \frac {\mathrm dz} {\mathrm dy} = \frac 2 y
So that:
\displaystyle \frac 1 {y^2} \left(\frac {\mathrm dy} {\mathrm dx}\right)^2 = 2 \ln y + C
At x = 0 we have y = 1 and \dfrac {\mathrm dy} {\mathrm dx} = 0, so C = 0. We therefore have:
\displaystyle \frac 1 y \left(\frac {\mathrm dy} {\mathrm dx}\right) = \sqrt {2 \ln y}
We therefore have:
\displaystyle \frac 1 {y \sqrt {\ln y}} \frac {\mathrm dy} {\mathrm dx} = \sqrt 2
Note that the LHS can be integrated by inspection or setting y = e^u, then:
\begin{align*}\int \frac 1 {y \sqrt {\ln y}} & = \int \frac {e^u} {e^u \sqrt u} \mathrm du \\ & = \int 1 {\sqrt u} \mathrm du \\ & = 2 \sqrt u + C \\ & = 2 \sqrt {\ln y} + C\end{align*}
So:
2 \sqrt {\ln y} = \sqrt 2 x + C
At x = 0 we have y = 1, so C = 0, giving:
2 \sqrt {\ln y} = \sqrt 2 x
Squaring:
4 \ln y = 2 x^2
So:
\ln y = \dfrac {x^2} 2
giving:
y = e^{x^2/2}
