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

[gikiian]

I have the ODE x''(t)+\frac{k}{m}x(t)=0.

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

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

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

Next, I check for the linear independence of the two solutions using the 2x2 Wronskian matrix and obtain Wronskian=-2ωsin(ø) 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?

 

[JJacquelin]

The two solutions are not necessarily with the same ø
x_{1}(t)=Asin(\sqrt\frac{k}{m}t+ø_1), x_{2}(t)=Bsin(-\sqrt\frac{k}{m}t+ø_2)

 

[pasmith]

I have the ODE x''(t)+\frac{k}{m}x(t)=0.

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

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

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

Next, I check for the linear independence of the two solutions using the 2x2 Wronskian matrix and obtain Wronskian=-2ωsin(ø) 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?

The general solution of x'' + \omega^2x =0 is
x = A\cos (\omega t)+ B\sin (\omega t)
By use of the sum formulae for sine and cosine this can be written as either
<br /> x = R \cos (\omega t + \phi)<br />
where R^2 = A^2 + B^2, \cos \phi = A and \sin\phi = -B, or as
<br /> x = R \sin (\omega t + \theta)<br />
where R^2 = A^2 + B^2, \cos \theta = B and \sin\theta = A.

The point is that the general solution has two parameters: either two amplitudes or an amplitude and a phase shift.

 

[PeroK]

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

<br /> <br /> 1) If two functions are linearly dependent, then the Wronskian will vanish everywhere. <br /> <br /> 2) If the Wronskian is not zero at any point, then the functions must be linearly independent.<br /> <br /> So, your two functions are linearly independent unless w = 0 or ø is a multiple of pi.