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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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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.
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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