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paccali
paccali is offline
#3
Oct8-10, 10:26 PM
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:

[tex]\frac{dz_{1}}{dt}=-z_{1}^{2}[/tex]
[tex]\frac{dz_{1}}{-z_{1}^{2}}=dt[/tex]

I then integrated each side:

[tex]\int -\frac{1}{z_{1}^{2}}dz=\int dt[/tex]
[tex]\frac{1}{z_{1}}+C_{1}=t+C_{2}[/tex]

Since both sides had constants, I dropped one, and I then used the initial condition of [tex]z_{1}(t=0)=x_{1}[/tex] to solve for C

[tex]\frac{1}{z_{1}}+C=t[/tex]
[tex]\frac{1}{z_{1}}=t-C[/tex]
[tex]z_{1}=\frac{1}{t-C}[/tex]
[tex]z_{1}(0)=x_{1}=\frac{1}{0-C}[/tex]
[tex]C=-\frac{1}{x_{1}}[/tex]

I plugged in C above and got this equation for Lagrangian position:

[tex]z_{1}=\frac{1}{t+\frac{1}{x_{1}}}=\frac{x_{1}}{1+tx_{1}}[/tex]