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
$$x = R \cos (\omega t + \phi)$$
where $R^2 = A^2 + B^2$, $\cos \phi = A$ and $\sin\phi = -B$, or as
$$x = R \sin (\omega t + \theta)$$
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 [itex] <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.[/itex]

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