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]

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.

1) If two functions are linearly dependent, then the Wronskian will vanish everywhere.

2) If the Wronskian is not zero at any point, then the functions must be linearly independent.

So, your two functions are linearly independent unless w = 0 or ø is a multiple of pi.

 

[blue_raver22]

To find the second solution of the ODE, you can use the method of reduction of order. This involves assuming a second solution of the form x_2(t)=v(t)x_1(t), where v(t) is a function to be determined. Substituting this into the ODE and solving for v(t) will give you the second solution. Alternatively, you can also use the method of variation of parameters to find the second solution. This involves finding a particular solution of the ODE and then using it to construct the second solution.