STEP II 1992 question 2 solution

\frac{\text{d}y}{\text{d}x} = x\frac{\text{d}^2y}{\text{d}x^2} + \frac{\text{d}y}{\text{d}x} - \frac{\text{d}^2y}{\text{d}x^2}\sinh \bigg(\frac{\text{d}y}{\text{d}x}\bigg)\\
0 = \frac{\text{d}^2y}{\text{d}x^2} \bigg( x - \sinh \bigg(\frac{\text{d}y}{\text{d}x}\bigg)\bigg) \\
\therefore \frac{\text{d}^2y}{\text{d}x^2} = 0 \Rightarrow y = Ax + B\\
\text{or } x - \sinh \bigg(\frac{\text{d}y}{\text{d}x}\bigg) = 0 \Rightarrow \sinh^{-1} x = \frac{\text{d}y}{\text{d}x} \\

\frac{\text{d}y}{\text{d}x} = \sinh^{-1} x + \frac{x}{\sqrt{1+x^2}} - \frac{x}{\sqrt{1+x^2}} = \sinh^{-1} x.\\
(*): x\sinh^{-1} x - (1+x^2)^{1/2} + C = x\sinh^{-1} x - \cosh (\sinh^{-1} x).\\
- (1+x^2)^{1/2} + C = - \cosh (\sinh^{-1} x).\\
u = \sinh^{-1} x \Rightarrow \sinh u = x \Rightarrow \cosh u = \sqrt{1+x^2} \\
\Rightarrow \cosh (\sinh^{-1} x) = (1+x^2)^{1/2}.\\
(*): - (1-x^2)^{1/2} + C = - (1+x^2)^{1/2} \\
C = 0\\