very differential equation with not known source function

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logistic_guy

Junior Member
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Apr 17, 2024
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here is the question

Solve the classical differential equation \(\displaystyle u'' - k^2u = f(x), \ \ \ \ a < x < b, \ \ \ \ u(a) = u(b) = 0\).

Hint: \(\displaystyle u(x) = \int_{a}^{b} G(x,s) f(s) \ ds\),

where the function \(\displaystyle G(x,s)\) is known as Green's function for the problem.

my attemp
before and on paper i find \(\displaystyle u(x) = c_1e^{kx} + c_2e^{-kx}\)

how to find particular solution for \(\displaystyle f(x)\)?
 
logistic_guy said:
i can't find it


can you help?
Click to expand...
Anyway, my hint was to use the Dirac delta function which means to solve this differential equation:

[imath]\displaystyle u'' - k^2u = \delta(x-s)[/imath]

The solution to this differential equation is [imath]\displaystyle G(x,s)[/imath].

Another way is to solve by using variation of parameters.

Do you know how to use these methods? Have you tried anyone of them?
 
logistic_guy said:
here is the question

Solve the classical differential equation \(\displaystyle u'' - k^2u = f(x), \ \ \ \ a < x < b, \ \ \ \ u(a) = u(b) = 0\).

Hint: \(\displaystyle u(x) = \int_{a}^{b} G(x,s) f(s) \ ds\),

where the function \(\displaystyle G(x,s)\) is known as Green's function for the problem.

my attemp
before and on paper i find \(\displaystyle u(x) = c_1e^{kx} + c_2e^{-kx}\)

how to find particular solution for \(\displaystyle f(x)\)?
Click to expand...
i insist to solve this differential. if i can't i'll quit engineering

mario99 said:
Do you know how to use these methods? Have you tried anyone of them?
Click to expand...
no:(

i know how to solve
\(\displaystyle u'' - k^2u = 0\)
\(\displaystyle u'' - k^2u = 1\)
\(\displaystyle u'' - k^2u = x\)
\(\displaystyle u'' - k^2u = x^2 + e^x\)
\(\displaystyle u'' - k^2u = \tan x\)
\(\displaystyle u'' - k^2u = x\sin x\)
 
logistic_guy said:
i insist to solve this differential. if i can't i'll quit engineering
Click to expand...
Then, solve it! What are you waiting for?

🤔

logistic_guy said:
i know how to solve
\(\displaystyle u'' - k^2u = 0\)
\(\displaystyle u'' - k^2u = 1\)
\(\displaystyle u'' - k^2u = x\)
\(\displaystyle u'' - k^2u = x^2 + e^x\)
\(\displaystyle u'' - k^2u = \tan x\)
\(\displaystyle u'' - k^2u = x\sin x\)
Click to expand...
I am better than you. I know how to solve:

[imath]\displaystyle u'' - k^2u = 0[/imath]
[imath]\displaystyle u'' - k^2u = 1[/imath]
[imath]\displaystyle u'' - k^2u = x[/imath]
[imath]\displaystyle u'' - k^2u = x^2 + e^x[/imath]
[imath]\displaystyle u'' - k^2u = \tan x[/imath]
[imath]\displaystyle u'' - k^2u = x\sin x[/imath]
[imath]\displaystyle u'' - k^2u = f(x)[/imath]

🥴
 
mario99 said:
Then, solve it! What are you waiting for?

🤔
Click to expand...
i can't i don't know how to solve it

mario99 said:
I am better than you. I know how to solve:

[imath]\displaystyle u'' - k^2u = 0[/imath]
[imath]\displaystyle u'' - k^2u = 1[/imath]
[imath]\displaystyle u'' - k^2u = x[/imath]
[imath]\displaystyle u'' - k^2u = x^2 + e^x[/imath]
[imath]\displaystyle u'' - k^2u = \tan x[/imath]
[imath]\displaystyle u'' - k^2u = x\sin x[/imath]
[imath]\displaystyle u'' - k^2u = f(x)[/imath]

🥴
Click to expand...
i need the solution of the last one
 
logistic_guy said:
i can't i don't know how to solve it


i need the solution of the last one
Click to expand...
Did you read about my suggestion in post #7? If no, go and read about how to solve differential equations by Dirac delta function or by variation of parameters. If you don't know how to use either of these methods, it will be pointless (And waste of time) to explain the solution of the op differential equation because eventually you will give up as what happened in the complex analysis problem.

Have you at least looked at the solution in W|A?
 
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