If dy dx= 2^xy+2^y· 2^x 2^x+2^x+y log_e2,y(0)=0, then for y=1, the value of x lies in the interval :

Given differential equation is

$$\frac{dy}{dx}=\frac{{2}^{x}\cdot y+{2}^{y}\cdot {2}^{x}}{{2}^{x}+{2}^{x+y}{\mathrm{log}}_{e}2}$$dxdy​=2x+2x+yloge​22x⋅y+2y⋅2x​

$$\Rightarrow \frac{dy}{dx}=\frac{{2}^{x}(y+{2}^{y})}{{2}^{x}(1+{2}^{y}{\mathrm{log}}_{e}2)}$$⇒dxdy​=2x(1+2yloge​2)2x(y+2y)​

$$\Rightarrow \int \frac{1+{2}^{y}{\mathrm{log}}_{e}2}{y+{2}^{y}}dy=\int dx$$⇒∫y+2y1+2yloge​2​dy=∫dx

$$\Rightarrow \int \frac{d(y+{2}^{y})}{y+{2}^{y}}=\int dx$$⇒∫y+2yd(y+2y)​=∫dx

$$\Rightarrow \mathrm{ln}|y+{2}^{y}|=x+C[\because \int \frac{{f}^{'}(x)}{f(x)}dx=\mathrm{ln}(f(x))+C]$$⇒ln∣y+2y∣=x+C[∵∫f(x)f′(x)​dx=ln(f(x))+C]

Now $\because y(0)=0$∵y(0)=0

$$\Rightarrow C=0$$⇒C=0

$$\therefore \mathrm{ln}|y+{2}^{y}|=x$$∴ln∣y+2y∣=x

Now for $y=1$y=1 we have

$$x=\mathrm{ln}(1+2)=\mathrm{ln}3\in (1,2)$$x=ln(1+2)=ln3∈(1,2)