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dantistus

Member

Registered: 2006-04-16

Posts: 2

Duamel formulae

My greetings Guys, my English may be not perfect, excuse me for that. I've got a problem when applying Duamel formulae to solve the following differential equation:

y''-y=1/cosh(t)^2

(Here y'' is twice diff'ed function y(t)). In order to apply Duamel formulae, I solve additional equation

Y''-Y=1

The solution of this equation it

Y=e^t-1

Then, I want to apply Duamel formulae, but everytime I get a wrong result The Duamel formulae I use is

y=integral (from 0 to t) (Y'(t-x)*f(x)dx)

Where Y' is differential of Y (Y'=e^t), f(x) is right part of given differential equation, e.g. 1/cosh(x)^2

Guys, I would be very thankful if you guide me to the solution of this eq Thank you

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HallsofIvy

Guest

Re: Duamel formulae

My greetings Guys, my English may be not perfect, excuse me for that. I've got a problem when applying Duamel formulae to solve the following differential equation:

y''-y=1/cosh(t)^2

(Here y'' is twice diff'ed function y(t)). In order to apply Duamel formulae, I solve additional equation

Y''-Y=1

The solution of this equation it

Y=e^t-1

Then, I want to apply Duamel formulae, but everytime I get a wrong result The Duamel formulae I use is

y=integral (from 0 to t) (Y'(t-x)*f(x)dx)

Where Y' is differential of Y (Y'=e^t), f(x) is right part of given differential equation, e.g. 1/cosh(x)^2

Guys, I would be very thankful if you guide me to the solution of this eq Thank you

So you want to integrate e^(t-x)/cosh^2 x.  It might be best to write cosh x in terms of the exponential: cosh x= (e^x- e^(-x))/2 so cosh^2 x= (e^(2x)- 1+ e^(-2x))/4.  Written like that you are integrating 4e^x/(e^(x))^2- 1+ (e^(x)^(-2)).  I would suggest the substitution u= e^x.

dantistus

Member

Registered: 2006-04-16

Posts: 2

Re: Duamel formulae

So you want to integrate e^(t-x)/cosh^2 x.  It might be best to write cosh x in terms of the exponential: cosh x= (e^x- e^(-x))/2 so cosh^2 x= (e^(2x)- 1+ e^(-2x))/4.  Written like that you are integrating 4e^x/(e^(x))^2- 1+ (e^(x)^(-2)).  I would suggest the substitution u= e^x.

Hey, thank you for the reply very much Unfortunately this want help me - this is exactly what I am doing. You've mistaken a little, cosh(x)=(e^x+e^(-x))/2, and cosh(x)^2= (e^(2x) + 2 + e^(-2x))/4 (as (a+b)^2 = a^2 + 2ab + b^2). Then, I need to integrate the following expression:

e^t*e^(-x)*4/(e^(2x) + 2 + e^(-2x))

I apply substitution proposed by you:

u=e^x
du=e^xdx

I don't have my papers with me, so I can't tell you what is my result for the integral, but it is wrong, e.g. with this solution

y''-y!=1/cosh(x)^2

And I do not know what to do, cause all my calculations seem to be proper, but the result is not the solution of the equation

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