Paper 2, Section I, A

Differential Equations | Part IA, 2013

Use the transformation z=lnxz=\ln x to solve

z¨=z˙21ez\ddot{z}=-\dot{z}^{2}-1-e^{-z}

subject to the conditions z=0z=0 and z˙=V\dot{z}=V at t=0t=0, where VV is a positive constant.

Show that when z˙(t)=0\dot{z}(t)=0

z=ln(V2+41)z=\ln \left(\sqrt{V^{2}+4}-1\right)

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