Draw a phase-plane direction field for the equation Hence sketch the solution path of the equation that
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Draw a phase-plane direction field for the equation
Hence sketch the solution path of the equation that starts from the initial conditions x = 1, dx/dt = 0.
Transcribed Image Text:
dx dt +1.5 d.x dt + 40x = 0 (10.60)
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Equation 1060 can be expressed as du dx so the curve on which the solutionpath gradient is equal to k is given by kv 15v Thus as shown in Figure 1036 the curves of constant solutionpath gradient are in this case cubic functions of v The solution ...View the full answer
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