dsolve unable to find a solution

2018 4 8
3 个回答
回答已采纳
更新时间：2018 4 8
5 次查看（30 天）

0 个投票

Hi, I need to find a solution for the differential equation (image attached), where theta, gamma and M are parameters. I am trying to solve the equations using dsolve (code attached). It returns an error "Warning: Unable to find explicit solution." Am I doing something wrong or should I use a different ode solver for this equation? Please help! Thanks!

请先登录，再回答此问题。

 采纳的回答

Sam K
Sam K 2018-4-8

0 个投票

Sorry about the confusion about the notation. The superscript "T" in my model is not the transpose. The notation denotes the type of source in my model, which is irrelevant to my question here. So, please ignore the superscript "T".

2 个评论
显示 无 隐藏 无

Walter Roberson
Walter Roberson 2018-4-8
In any case you have QT multiplied by something in the image but evaluated at something instead of multiplied in the code.
Walter Roberson
Walter Roberson 2018-4-8
>> syms theta lambda M
>> a = (1 - theta/exp(lambda*M));
>> syms Q(t)
>> eqn = diff(Q,t) == Q*a;
cond = Q(0) == 400;
qSol(t) = dsolve(eqn,cond)
qSol(t) =
400*exp(-t*exp(-M*lambda)*(theta - exp(M*lambda)))

请先登录，再进行评论。

更多回答（2 个）

Walter Roberson
Walter Roberson 2018-4-8

0 个投票

Your image shows Q transpose multiplied by something, but your function has Q evaluated at something not multiplied.
Sam K
Sam K 2018-4-8
编辑：Sam K 2018-4-8

0 个投票

Hi walter,
Thank you for helping me out with this. I see your point. I guess that is part of my confusion on how to write the code. My equation for Q, as shown in the picture, is: dQ/dt = Q*a, where a is the parameters in the parentheses. I need to plot this equation based on the initial value of Q. I thought dsolve would be a good solver, but I must be doing something wrong with the code since I am a newbie to Matlab coding. Should I change the code as follows:
a = (1 - theta/exp(lambda*M));
syms Q(t) a
eqn = diff(Q,t) == Q*a;
cond = Q(0) == 400;
qSol(t) = dsolve(eqn,cond);

类别

在 帮助中心 和 File Exchange 中查找有关 Numeric Solvers 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by