JEE Main 2020 — Mathematics Calculus

$$(1+{e}^{-x})(1+{y}^{2})\frac{dy}{dx}={y}^{2}$$(1+e−x)(1+y2)dxdy​=y2

Separate variables

$$\Rightarrow \frac{(1+{y}^{2})dy}{{y}^{2}}=\frac{dx}{1+{e}^{-x}}$$⇒y2(1+y2)dy​=1+e−xdx​

Integrate both side

$$\Rightarrow \int \frac{(1+{y}^{2})dy}{{y}^{2}}=\int \frac{dx}{1+{e}^{-x}}$$⇒∫y2(1+y2)dy​=∫1+e−xdx​

$$\Rightarrow \int (1+\frac{1}{{y}^{2}})dy=\int \frac{{e}^{x}dx}{{e}^{x}+1}$$⇒∫(1+y21​)dy=∫ex+1exdx​

$$\Rightarrow y-\frac{1}{y}=\mathrm{ln}({e}^{x}+1)+C$$⇒y−y1​=ln(ex+1)+C

Since curve pass through $(0,1)$(0,1).

$$\Rightarrow 1-\frac{1}{1}=\mathrm{ln}({e}^{0}+1)+C\Rightarrow C=-\mathrm{ln}(2)$$⇒1−11​=ln(e0+1)+C⇒C=−ln(2)

$$\Rightarrow y-\frac{1}{y}=\mathrm{ln}({e}^{x}+1)-\mathrm{ln}(2)$$⇒y−y1​=ln(ex+1)−ln(2)

$$\Rightarrow \frac{{y}^{2}-1}{y}=\mathrm{ln}(\frac{{e}^{x}+1}{2})$$⇒yy2−1​=ln(2ex+1​)

$$\Rightarrow {y}^{2}=1+y{\mathrm{log}}_{e}(\frac{1+{e}^{x}}{2})$$⇒y2=1+yloge​(21+ex​)