Integral simplification with dummy variable s

[Xyius]

Integral simplification with dummy variable "s"

I am looking at the derivation for the solution to solve the partial differential equation of a wave with a time-dependent forcing term h(x,t). In the derivation they get to a point where they need to solve the following second order ODE.

$$u''(t)+(\frac{\alpha n \pi}{L})^2u(t)=h(t)$$

The method here would be variation of parameters. I obtained a homogeneous solution of..

$$c_1cos(\frac{\alpha n \pi}{L}t)+c_2sin(\frac{\alpha n \pi}{L}t)$$
Which the book also got.
I got this for the particular solution after using variation of parameters...
$$\int h(t)cos(\frac{\alpha n \pi}{L}t)dt-\int h(t)sin(\frac{\alpha n \pi}{L}t)dt$$

The book got..
$$\frac{L}{n \pi \alpha} \int^{t}_{0} h(s)sin[\frac{\alpha n \pi}{L}(t-s)]ds$$

Now I know these must somehow be the same (unless I messed up using variation of parameters) but I am new to using a dummy variable "s" to simplify integrals like this. I have seen it a couple times but do not have much practice with it. If anyone could help me through this it would be great!

Thanks!

 

[LCKurtz]

I am looking at the derivation for the solution to solve the partial differential equation of a wave with a time-dependent forcing term h(x,t). In the derivation they get to a point where they need to solve the following second order ODE.

$$u''(t)+(\frac{\alpha n \pi}{L})^2u(t)=h(t)$$

The method here would be variation of parameters. I obtained a homogeneous solution of..

$$c_1cos(\frac{\alpha n \pi}{L}t)+c_2sin(\frac{\alpha n \pi}{L}t)$$
Which the book also got.
I got this for the particular solution after using variation of parameters...
$$\int h(t)cos(\frac{\alpha n \pi}{L}t)dt-\int h(t)sin(\frac{\alpha n \pi}{L}t)dt$$

The book got..
$$\frac{L}{n \pi \alpha} \int^{t}_{0} h(s)sin[\frac{\alpha n \pi}{L}(t-s)]ds$$

Now I know these must somehow be the same (unless I messed up using variation of parameters) but I am new to using a dummy variable "s" to simplify integrals like this. I have seen it a couple times but do not have much practice with it. If anyone could help me through this it would be great!

Thanks!

I will use y for your dependent variable and rename your constant

$$k =\frac{\alpha n \pi}{L}$$

$$y'' + k^2 y = h(t)$$

and your homogeneous solution pair {sin(kt), cos(kt)}. OK so far. Now you looked for a particular solution

y_p = u cos(kt) + v sin(kt)

and you should have gotten (I think you missed a k in the denominator which you can fix):

$$u' = -\frac{ h(t)\sin(kt)}{k},\ v' = \frac {h(t)\cos(kt)}{k}$$

Now integrating to find u and v is where you can express the integrals as functions of the upper limit with a definite integral using a dummy variable for the integrand. Using 0 for the lower limit just affects the constant of integration but you are looking for any particular solution so you don't care. You could use any lower limit:

$$u(t) = \int_0^t -\frac{ h(s)\sin(ks)}{k}\, ds$$

$$v(t) = \int_0^t \frac {h(s)\cos(ks)}{k}\, ds$$

Now, remembering that y_p = u cos(kt) + v sin(kt) you get

$$y_p = \cos(kt)\int_0^t -\frac{ h(s)\sin(ks)}{k}\, ds + \sin(kt)\int_0^t \frac {h(s)\cos(ks)}{k}\, ds$$

Now you can rewrite that as one big integral with the 1/k out in front and the sin(kt) and cos(kt) inside the integrals because they don't depend on s and you will see an addition formula you should recognize, not to mention the book's answer.

 

[Xyius]

Ohh! Thank you very much! I understand now :D