JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Differential Equations

Separate the variables:



With $f(0)=0$: $\frac{1}{20}\ln 1=c$, so $c=0$ and



At $x=0$, $y=0$ and $\frac{5y-2}{5y+2}=-1<0$; the solution stays in $-\frac{2}{5}<y<\frac{2}{5}$ (it can never cross the constant solutions $y=\pm \frac{2}{5}$), so the ratio stays negative and $\frac{2-5y}{2+5y}=e^{20x}$. Solving for $y$:



As $x\to -\infty$, $e^{20x}\to 0$, so



(Check: at $y=0$, $\frac{dy}{dx}=-4<0$, so $f$ decreases as $x$ increases — it climbs towards $\frac{2}{5}$ going left and falls towards $-\frac{2}{5}$ going right.)

Answer: 0.40