Given that
\[y = A \cos [\ln(x+1)] + B \sin [\ln(x+1)],\]
where&nbsp;\(A\) and \(B\) are constants with \(x &gt; -1\).
Show that&nbsp;\((x+1)^2 \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + (x+1) \frac{\mathrm{d}y}{\mathrm{d}x} + y = 0\).[5]

\[y = A \cos [\ln(x+1)] + B \sin [\ln(x+1)],\]

where \(A\) and \(B\) are constants with \(x > -1\).

Show that \((x+1)^2 \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + (x+1) \frac{\mathrm{d}y}{\mathrm{d}x} + y = 0\).

2018 AJC C1BT1 Q3

2018 AJC C1BT1 Q3

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