STEP III 2012 1

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}

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