If dydx=2^xy+2y2x^22^x+2x+y_e2 (schematic), y(0)=0, then for y=1, x lies in:

Question

If $\dfrac{dy}{dx}=\dfrac{2^{xy+2y}\cdot2x^2}{2^{x+2x+y}}\log_e2\ (	ext{schematic})$, $y(0)=0$, then for $y=1$, $x$ lies in:

Accepted Answer

Simplifying to $\dfrac{dy}{dx}=\dfrac{y+2^y}{1+2^y}\ln2$, separating and integrating gives $\log|y+2^y|=x+C$; with $y(0)=0 \Rightarrow C=0$.

For $y=1$: $x=\ln|1+2|=\ln3\approx1.1 \implies x\in(1,2)$.

Solutions · 1