A problem about laplace

[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 text book 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.