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

• gikiian
In summary, the given ODE has two solutions x_{1}(t)=Asin(\sqrt\frac{k}{m}t+ø) and x_{2}(t)=Bsin(-\sqrt\frac{k}{m}t+ø), which are not necessarily with the same ø. The general solution of the ODE is x = R \cos (\omega t + \phi) or x = R \sin (\omega t + \theta), where R^2 = A^2 + B^2 and either \cos \phi = A, \sin\phi = -B or \cos \theta = B, \sin\theta = A. The Wronskian of the two solutions is -2
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?

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

gikiian said:
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.

gikiian said:
I have the ODE $Next, I check for the linear independence of the two solutions using the 2x2 Wronskian matrix and obtain [itex]Wronskian=-2ωsin(ø)$ 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.

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.

## 1. What is the meaning of the equation "Solving x''(t)+Kx(t)=0 when x has the form sin(ωt+ø): the soln r LD"?

The equation represents a differential equation where the second derivative of the function x with respect to time (t) plus the product of the function x and the constant K is equal to zero. The solution to this equation is a sinusoidal function with a constant amplitude (r) and a constant phase difference (ø) with respect to the angular frequency (ωt).

## 2. How do you solve for x in this equation?

To solve for x, we can use the standard method for solving linear differential equations. First, we assume that the solution has the form x(t) = r sin(ωt+ø). Then, we substitute this into the original equation and solve for the constants r and ø. Finally, we can plug in these values into our solution to get the final answer of x(t) = r sin(ωt+ø).

## 3. What is the significance of the constants r and ø in the solution?

The constant r represents the amplitude of the sinusoidal function, while the constant ø represents the phase difference with respect to the angular frequency. These constants determine the shape and position of the graph of the solution function.

## 4. Are there any restrictions on the values of the constants in this equation?

Yes, there are certain restrictions on the values of the constants to ensure that the solution remains a sinusoidal function. The amplitude (r) must be a positive value, the phase difference (ø) must be within the range of -π to π, and the angular frequency (ω) must be a non-zero value.

## 5. Can this equation be applied to other types of differential equations?

Yes, this equation can be applied to other types of linear differential equations with different forms for the function x(t). As long as the equation has the form of x''(t)+Kx(t)=0, the solution can be expressed as a sinusoidal function with the constants r and ø. This method is commonly used in physics and engineering to solve differential equations in oscillatory systems.

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