I have the ODE [itex]x''(t)+\frac{k}{m}x(t)=0[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

Given that the solution is of the form [itex]sin(ωt+ø)[/itex], I plug this form into the original ODE and obtain [itex]ω=+\sqrt\frac{k}{m},-\sqrt\frac{k}{m}[/itex].

And hence, I obtain two solutions of the ODE as follows:

[itex]x_{1}(t)=Asin(\sqrt\frac{k}{m}t+ø), x_{2}(t)=Bsin(-\sqrt\frac{k}{m}t+ø)[/itex]

Next, I check for the linear independence of the two solutions using the 2x2 Wronskian matrix and obtain [itex]Wronskian=-2ωsin(ø)[/itex] which vanishes at ø=0 or ω=0, implying that the two solutions are not linearly independent.

How can I find the second solution of the second order ODE?

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# Solving x''(t)+Kx(t)=0 when x has the form sin(ωt+ø): the soln r LD!

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