Can Laplace Transform Solve Differential Equations with Arbitrary Constants?

[bfaskiplar]

hi guys

i cannot solve this.

say that y'' + y = g(t) derive the formula y(t) = c1*cost + c2*sint + convolution of sint and g(t)

 

[jackmell]

Either solve it via variation of parameters or since you mentioned laplace, take the laplace transform of both sides:

$$\mathcal{L}\left\{y''+y=f(x)\right\}$$

and letting:

$$\mathcal{L}\left\{y\right\}=\widetilde{y}$$

solve for $\widetilde{y}$, invert, and use the convolution theorem to express the solution in terms of a convolution.

Look in any DE textbook and this problem will be solved both ways.

 

[Dickfore]

Use the method of variation of arbitrary constants. The solution to the homogeneous equation is:

$$y_{0}(t) = C_{1} \, \cos{t} + C_{2} \, \sin{t}$$

Then, assume $C_{i} \rightarrow C_{i}(t), i = 1, 2$ where these functions satisfy the following conditions:

$$\left[\begin{array}{cc} \cos{t} & \sin{t} \\ -\sin{t} & \cos{t} \end{array}\right] \cdot \left[\begin{array}{c} C'_{1}(t) \\ C'_{2}(t) \end{array}\right] = \left[\begin{array}{c} 0 \\ g(t) \end{array}\right]$$

The solution for $C'_{i}(t)$ is:

$$\left[\begin{array}{c} C'_{1}(t) \\ C'_{2}(t) \end{array}\right] = \left[\begin{array}{cc} \cos{t} & -\sin{t} \\ \sin{t} & \cos{t} \end{array}\right] \cdot \left[\begin{array}{c} 0 \\ g(t) \end{array}\right] = \left[\begin{array}{c} -\sin{t} \, g(t) \\ \cos{t} \, g(t) \end{array}\right]$$

One integration gives the following:

$$C_{1}(t) = C_{1} - \int_{t_{0}}^{t}{g(t') \, \sin{t'} \, dt'}$$

$$C_{2}(t) = C_{2} + \int_{t_{0}}^{t}{g(t') \, \cos{t'} \, dt'}$$

Substituting this into the expression for the general solution, one gets:

$$y(t) = C_{1} \, \cos{t} + C_{2} \, \sin{t} + \int_{t_{0}}^{t}{g(t') \, \left(\sin{t} \, \cos{t'} - \cos{t} \, \sin{t'}\right) \, dt' \right)}$$

where, the integrating constants $C_{1/2}$, are determined from the initial conditions:

$$y(t_{0}) = C_{1}$$

$$y'(t_{0}) = C_{2}$$

Using the addition theorem for the sine function:

$$\sin{(t - t')} = \sin{t} \, \cos{t'} - \cos{t} \, \sin{t'}$$

we see that the above integral can be written as:

$$\int_{t_{0}}^{t}{g(t') \, \sin{(t - t')} \, dt'}$$

Take $t_{0} = 0$ and compare with the definition for convolution, you will get your desired result.