Given that $$y = A \cos [\ln(x+1)] + B \sin [\ln(x+1)],$$ where $A$ and $B$ are constants w

Question

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

Accepted Answer

$(x+1)^2\dfrac{\mathrm d^2y}{\mathrm dx^2}+(x+1)\dfrac{\mathrm dy}{\mathrm dx}+y=0$

2018 AJC C1BT1 Q3

2018 AJC C1BT1 Q3

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