@Mark : What other way can he use to prove that 's' exists besides integration? It is given that after proving that s(x) exists, to show that it is bounded.
The integral isn't an elementary one though...can't be done using Riemann integrals.
Did they say that you can't use the variable-separable method after substitution? I think you have to apply the substitution and then use the variable-separable method...meaning all u terms on the LHS.
You needn't worry about 'n'. It isn't being integrated. Worry about 't', and 'n' will resolve itself after a little algebra. In simpler terms, n is a constant and t is the variable during integration.
and the formulae here -:
http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities
will help you in solving the integral ;). Though I don't know why a simple by parts integration doesn't suit you..
For e^(2t), the coefficient would be 2. e^(2t) won't make it zero, so just put 2 into D^2 + 8D and you get the particular integral as e^(2t) * 22/20. However, in 95e^(0t), it will be zero, so differentiate till it will be non zero on putting D = 0, OR until you get a constant after...
Wait a minute...the second last integral is first done wrt y and then the last integral is done wrt x. How did you get cos(x²)? cos(x) would be a constant wrt y. After correction, the last integral will simply become a by parts integral.
Well, how I've been taught involves finding the complementary function and the particular integral and adding them to get the complete solution. I'd follow these steps -:
1. Write in terms of the D operator. Here it would be -: (D² + 8D)y = 22exp(t) + 95exp(0).
I wrote 95 as 95exp(0) so that I...
To be precise, to complete the square follow these steps -:
1. Take the coefficient of the x term, half it.
2. Square this halved coefficient and then add and subtract it to the expression.
3. Complete the square.
Surely you can finish this now.
Mate, you need to use partial fractions to separate the denominator and integrate on each term separately.
Find A and B using the concept of partial fractions (this page is pretty neat...teaches you everything you need to know about them ...