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I am working through the following paper: https://www.researchgate.net/publication/222417459_Mathematical_models_for_motion_of_the_rear_ends_of_vehicles

On Page 4, equation 8 models the position of the back wheel when given the position of the front wheels during a turn. I wanted to model the path of the front and rear wheels when a car turns in a circle, which yielded the following differential equation:

$$\frac{d\psi}{dt}=\frac{\sin{\psi}\cos{t}-\cos{\psi}\sin{t}}{2.5}$$

How would I approach solving this equation. I have considered the following:

  • separating the variables is impossible
  • it isn't a linear first order differential equation so Euler's method of using an integrating factor will not work
  • I cannot see how to get $\frac{y}{x}$ used to solve a homogeneous differential equation.

Any help would be appreciated.

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$$\frac{d\psi}{dt}=\alpha\:\left(\sin{\psi}\cos{t}-\cos{\psi}\sin{t}\right)\quad;\quad \alpha=1/2.5$$ $$\frac{d\psi}{dt}=\alpha\:\sin(\psi-t)$$ $\psi(t)=y(t)+t$ $$\frac{dy}{dt}+1=\alpha\:\sin(y)$$ $$dt=\frac{dy}{\alpha\:\sin(y)-1}$$ $$t=\int\frac{dy}{\alpha\:\sin(y)-1}+\text{constant}$$ $$t=\frac{2}{\sqrt{1-\alpha^2}}\tan^{-1}\left(\frac{1}{\sqrt{1-\alpha^2}}\left(\alpha-\tan(\frac{y}{2}) \right) \right)+c$$ $$\sqrt{1-\alpha^2}\tan\left(\frac{\sqrt{1-\alpha^2}}{2 }(t-c)\right)=\alpha-\tan(\frac{y}{2}) $$ $$y=2\tan^{-1}\left(\alpha-\sqrt{1-\alpha^2}\tan\left(\frac{\sqrt{1-\alpha^2}}{2 }(t-c)\right) \right)$$ $$\psi(t)=t+2\tan^{-1}\left(\alpha-\sqrt{1-\alpha^2}\tan\left(\frac{\sqrt{1-\alpha^2}}{2 }(t-c)\right) \right)$$

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  • $\begingroup$ Thank you. How did you integrate from line 5 to line 6? $\endgroup$ – George Tian Jul 29 '19 at 8:00
  • $\begingroup$ Change of variable $\tan(y/2)=x$. Compute $\sin(y)$ and $dy$ in terms of $x$ and $dx$. $\endgroup$ – JJacquelin Jul 29 '19 at 8:37

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