Chapter 1 - First-Order Differential Equations - 1.12 Chapter Review - Additional Problems - Page 110: 15

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Given: $$mv\frac{dv}{dy}=mg-kv^2$$ Let $v^2=u \rightarrow \frac{du}{dy}=2v\frac{dv}{dy}$ We will have: $\frac{m}{2}\frac{du}{dt}=mg-ku\\ \rightarrow u'+\frac{k}{m}u=g$ then $u=e^{-\int \frac{k}{m}dy}(c_1+\int e^{\int \frac{k}{m}dy}gdy)=c_1e^{-\frac{ky}{m}}+\frac{mg}{k}$ Since $u(0)=0 \rightarrow c_1=-\frac{mg}{k}$ Hence, $v^2=\frac{mg}{k}(1-e^{-\frac{ky}{m}})$