2nd order non-linear differential equation

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DJOULDE Aristide
DJOULDE Aristide 2023 年 8 月 22 日
編集済み: Florian Bidaud 2023 年 8 月 22 日
I have a heat transfer problem and I need to derive an 2nd order non-linear differential equation.
After replacing all the constants with A and B. My equation looks like this.
T'' + A(T)^4+ B= 0
Boundery conditions are T'(x=0)=C(T(0)-D), and T'(x=L)=E(D-T(L))
A,B,C,D,E are constants
And I need to solve for T(x).
Thank you in advance for your help.

回答 (1 件)

Florian Bidaud
Florian Bidaud 2023 年 8 月 22 日
編集済み: Florian Bidaud 2023 年 8 月 22 日
If you have symoblic toolbox,
That would give something like that :
syms y(t)
Dy = diff(y);
ode = diff(y,t,2) + A*y^4 + B == 0
cond1 = Dy(0) == C*(y(0)-D); % Correction made after Torsten comment T(0) --> y(0)
cond2 = Dy(L) == E*(D-y(L)); % Correction made after Torsten comment T(L) --> y(L)
conds = [cond1 cond2];
ySol(x) = dsolve(ode,conds);
ySol = simplify(ySol)
  2 件のコメント
Torsten
Torsten 2023 年 8 月 22 日
cond1 = Dy(0) == C*(y(0)-D);
cond2 = Dy(L) == E*(D-y(L));
ySol(t) = dsolve(ode,conds);
instead of
cond1 = Dy(0) == C*(T(0)-D);
cond2 = Dy(L) == E*(D-T(L));
ySol(x) = dsolve(ode,conds);
Florian Bidaud
Florian Bidaud 2023 年 8 月 22 日
@Torsten Indeed, thank you for spotting that

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