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Previous Year Question (PYQs)
4
Solution of the differential equation
$\displaystyle \frac{dx}{dy}-\frac{x\log x}{1+\log x}=\frac{e^{y}}{1+\log x},\ \text{ if } y(1)=0,$ is –
Solution
Treat $x$ as a function of $y$ and set $u=1+\log x\ (\Rightarrow x=e^{u-1},\ \frac{dx}{dy}=x\frac{du}{dy})$.
The DE becomes
$u\frac{du}{dy}=u-1+e^{\,1+y-u}.$
This transforms to a first-order non-linear equation in $u(y)$ whose
implicit integral does **not** reduce to any of the listed closed forms (A)–(C).
With the initial condition $y(1)=0$ (i.e., $u=1$ at $y=0$), the solution is an
implicit relation not matching (A)–(C).
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