STEP II 1992 question 2 solution

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STEP II 1992 question 2 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1992 question 2 solution

$\displaystyle y = x\frac{\text{d}y}{\text{d}x} - \cosh\bigg(\frac{\text{d}y}{\text{d}x}\bigg)\; (*)\\ \frac{\text{d}y}{\text{d}x} = x\frac{\text{d}^2y}{\text{d}x^2} + \frac{\text{d}y}{\text{d}x} - \frac{\text{d}^2y}{\text{d}x^2}\sinh \bigg(\frac{\text{d}y}{\text{d}x}\bigg)\\ 0 = \frac{\text{d}^2y}{\text{d}x^2} \bigg( x - \sinh \bigg(\frac{\text{d}y}{\text{d}x}\bigg)\bigg) \\ \therefore \frac{\text{d}^2y}{\text{d}x^2} = 0 \Rightarrow y = Ax + B\\ \text{or } x - \sinh \bigg(\frac{\text{d}y}{\text{d}x}\bigg) = 0 \Rightarrow \sinh^{-1} x = \frac{\text{d}y}{\text{d}x} \\ \Rightarrow y = \int \sinh^{-1} \text{d}x = x \sinh^{-1} x - \int\frac{x}{\sqrt{1 + x^2}} \text{d}x = x \sinh^{-1} x - (1+x^2)^{1/2} + C.$

$\displaystyle (1)\;\; y = Ax+B: \frac{\text{d} y}{\text{d} x} = A. \\ (*): Ax + B = Ax - \cosh A \\ B = -\cosh A \\ \therefore y = Ax - \cosh A.$

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