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:

<br /> y_{0}(t) = C_{1} \, \cos{t} + C_{2} \, \sin{t}<br />

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

<br /> \left[\begin{array}{cc}<br /> \cos{t} &amp; \sin{t} \\<br /> <br /> -\sin{t} &amp; \cos{t}<br /> \end{array}\right] \cdot \left[\begin{array}{c}<br /> C&#039;_{1}(t) \\<br /> <br /> C&#039;_{2}(t)<br /> \end{array}\right] = \left[\begin{array}{c}<br /> 0 \\<br /> g(t)<br /> \end{array}\right]<br />

The solution for C&#039;_{i}(t) is:

<br /> \left[\begin{array}{c}<br /> C&#039;_{1}(t) \\<br /> <br /> C&#039;_{2}(t)<br /> \end{array}\right] = \left[\begin{array}{cc}<br /> \cos{t} &amp; -\sin{t} \\<br /> <br /> \sin{t} &amp; \cos{t}<br /> \end{array}\right] \cdot \left[\begin{array}{c}<br /> 0 \\<br /> g(t)<br /> \end{array}\right] = \left[\begin{array}{c}<br /> -\sin{t} \, g(t) \\<br /> <br /> \cos{t} \, g(t)<br /> \end{array}\right]<br />

One integration gives the following:

<br /> C_{1}(t) = C_{1} - \int_{t_{0}}^{t}{g(t&#039;) \, \sin{t&#039;} \, dt&#039;}<br />

<br /> C_{2}(t) = C_{2} + \int_{t_{0}}^{t}{g(t&#039;) \, \cos{t&#039;} \, dt&#039;}<br />

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

<br /> y(t) = C_{1} \, \cos{t} + C_{2} \, \sin{t} + \int_{t_{0}}^{t}{g(t&#039;) \, \left(\sin{t} \, \cos{t&#039;} - \cos{t} \, \sin{t&#039;}\right) \, dt&#039; \right)}<br />

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

<br /> y(t_{0}) = C_{1}<br />

<br /> y&#039;(t_{0}) = C_{2}<br />

Using the addition theorem for the sine function:

<br /> \sin{(t - t&#039;)} = \sin{t} \, \cos{t&#039;} - \cos{t} \, \sin{t&#039;}<br />

we see that the above integral can be written as:

<br /> \int_{t_{0}}^{t}{g(t&#039;) \, \sin{(t - t&#039;)} \, dt&#039;}<br />

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