Question

$\frac{dy}{dx}=2e^x{y}^3,y\left(0\right)=\frac{1}{2}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}=2e^x{y}^3,y\left(0\right)=\frac{1}{2} \\ \Rightarrow \frac{1}{y^3}dy=2e^{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{y^3}dy=\int 2e^{x}dx \\ \Rightarrow -\frac{1}{2y^2}=2e^x+C.....\left(1\right) \\ \mathrm{Given}:\mathrm{at}x=0,y=\frac{1}{2} \\ \mathrm{Substituting}\mathrm{the}\mathrm{values}\mathrm{of}x\mathrm{and}y\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ -\frac{1}{2\times {\displaystyle \frac{1}{4}}}=2e^0+C \\ \Rightarrow C=-2-2 \\ \Rightarrow C=-4 \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{C}\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \Rightarrow -\frac{1}{2y^2}=2e^x-4 \\ \Rightarrow y^2\left(8-4e^x\right)=1 \\ \mathrm{Hence},y^2\left(8-4e^x\right)=1\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$