Answer:
[tex]y=-\ln\left(-\frac{(x+2)^3}{3}+10\right)[/tex]
Step-by-step explanation:
Using separation of variables we have [tex]y'=\frac{dy}{dx}=(x+2)^2e^y \Rightarrow e^{-y}dy=(x+2)^2dx \Rightarrow \int e^{-y}dy=\int(x+2)^2dx \Rightarrow - e^{-y}=\frac{(x+2)^3}{3}+k[/tex], using the initial condition [tex]y(1)=0[/tex], we obtain [tex]-e^{-0}=\frac{(1+2)^3}{3}+k \Rightarrow k=-e^{-0}-3^2=-1-9=-10[/tex] and clearing [tex]y[/tex] it follows [tex]-y=\ln e^{-y}=\ln \left(-\frac{(x+2)^3}{3}+10\right) \Rightarrow y=-\ln\left(-\frac{(x+2)^3}{3}+10\right)[/tex].
