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 P: 6 Nevermind, I figured out how to solve the problem using my old differential equations textbook. In case someone is curious, here's what I did: I rearranged the terms so that like terms were on the same side: $$\frac{dz_{1}}{dt}=-z_{1}^{2}$$ $$\frac{dz_{1}}{-z_{1}^{2}}=dt$$ I then integrated each side: $$\int -\frac{1}{z_{1}^{2}}dz=\int dt$$ $$\frac{1}{z_{1}}+C_{1}=t+C_{2}$$ Since both sides had constants, I dropped one, and I then used the initial condition of $$z_{1}(t=0)=x_{1}$$ to solve for C $$\frac{1}{z_{1}}+C=t$$ $$\frac{1}{z_{1}}=t-C$$ $$z_{1}=\frac{1}{t-C}$$ $$z_{1}(0)=x_{1}=\frac{1}{0-C}$$ $$C=-\frac{1}{x_{1}}$$ I plugged in C above and got this equation for Lagrangian position: $$z_{1}=\frac{1}{t+\frac{1}{x_{1}}}=\frac{x_{1}}{1+tx_{1}}$$