JEE Main 2021 — Mathematics Calculus

Given that

$$\frac{dy}{dx}=\frac{{2}^{x+y}-{2}^{x}}{{2}^{y}}={2}^{x}(\frac{{2}^{y}-1}{{2}^{y}})$$dxdy​=2y2x+y−2x​=2x(2y2y−1​)

$$\Rightarrow (\frac{{2}^{y}}{{2}^{y}-1})dy={2}^{x}dx$$⇒(2y−12y​)dy=2xdx

By using Variable Separable method

$$\int \frac{{2}^{y}dy}{{2}^{y}-1}=\int {2}^{x}dx$$∫2y−12ydy​=∫2xdx

Let ${2}^{y}-1=t\Rightarrow {2}^{y}\mathrm{ln}(2)dy=dt$2y−1=t⇒2yln(2)dy=dt

$$\int \frac{dt}{t}\frac{1}{\ell n2}=\int {2}^{x}dx$$∫tdt​ℓn21​=∫2xdx

We know that \displaystyle \int \frac{1}{x}dx=\mathrm{ln}(x)+c&\int {a}^{x}dx=\frac{{a}^{x}}{\mathrm{ln}(a)}+c.

$$\Rightarrow \frac{\mathrm{ln}(t)}{\mathrm{ln}(2)}=\frac{{2}^{x}}{\mathrm{ln}(2)}+\frac{c}{\mathrm{ln}(2)}$$⇒ln(2)ln(t)​=ln(2)2x​+ln(2)c​

$$\Rightarrow \mathrm{ln}(t)={2}^{x}+c$$⇒ln(t)=2x+c

$$\Rightarrow \mathrm{ln}({2}^{y}-1)={2}^{x}+c$$⇒ln(2y−1)=2x+c

when $x=0,y=1\Rightarrow \mathrm{ln}(2-1)={2}^{0}+c$x=0,y=1⇒ln(2−1)=20+c

$$\Rightarrow c=-1.$$⇒c=−1.

$$\Rightarrow \mathrm{ln}({2}^{y}-1)={2}^{x}-1$$⇒ln(2y−1)=2x−1

$$\Rightarrow {2}^{y}-1={e}^{{2}^{x}-1}$$⇒2y−1=e2x−1

$$\Rightarrow {2}^{y}=1+{e}^{{2}^{x}-1}$$⇒2y=1+e2x−1

when $x=1\Rightarrow {2}^{y}=1+{e}^{1}\Rightarrow y={\mathrm{log}}_{2}(1+e)$x=1⇒2y=1+e1⇒y=log2​(1+e)