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Differential equation of second order with two variables

์กฐํšŒ ์ˆ˜: 9 (์ตœ๊ทผ 30์ผ)
Florian Tahiraj
Florian Tahiraj 2023๋…„ 6์›” 5์ผ
๋‹ต๋ณ€: Pranjal Saxena 2023๋…„ 7์›” 28์ผ
to be solved is the equation ๐ธ๐ผ๐‘ค4(๐‘ฅ,๐‘ก)+๐‘š ๐‘คฬˆ(๐‘ฅ,๐‘ก)=0 and this can be solved by expressing ๐‘ค(๐‘ฅ,๐‘ก)= ๐œ™(๐‘ฅ)โ‹…๐‘Œ(๐‘ก)
for ๐œ™(๐‘ฅ) we do have this expression ๐œ™๐‘–(๐‘ฅ)=๐ดโ‹…[sin(๐œ†๐‘– ๐‘ฅ๐ฟ)โˆ’sinh(๐œ†๐‘– ๐‘ฅ๐ฟ)+sin(๐œ†๐‘–)+sinh(๐œ†๐‘–)cos(๐œ†๐‘–)+cosh(๐œ†๐‘–)โ‹…(cosh(๐œ†๐‘– ๐‘ฅ๐ฟ)โˆ’cos(๐œ†๐‘–๐‘ฅ๐ฟ)) ]
and for ๐‘Œ๐‘–(๐‘ก)=๐‘Œ(0)cos(๐œ”๐‘– ๐‘ก)+๐‘Œฬ‡(0)๐œ”sin(๐œ”๐‘–(๐‘ก))=๐ถ(0)โ‹…๐‘๐‘œ๐‘ (๐œ”๐‘– ๐‘ก+๐œƒ(0)). Whereas ๐œ†๐‘– is solved through this equation 1+cos(๐œ†๐‘–)โ‹…cosh(๐œ†๐‘–)=0
I am guessing that to solve the first differential equation as conditions we can use the three other equations given.
Does anyone has some tipps how this differential equation can be solved?

๋‹ต๋ณ€ (1๊ฐœ)

Pranjal Saxena
Pranjal Saxena 2023๋…„ 7์›” 28์ผ
Hi Florian,
I understand that you want to solve this second order differential equation.
You can use the โ€œSymbolic Math Toolboxโ€ in MATLAB to do so.
A typical approach to solving higher-order ordinary differential equations is to convert them to systems of first-order differential equations, and then solve those systems.
First you need to convert the second-order differential equation into a system of differential equations that can be solved using the numerical solver โ€œode45โ€ of MATLAB.
You can refer to the following MATLAB documentations for more information:
I hope this helps.
Warm Regards,
Pranjal.

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