开发者

Problem Solving Differential Equation Initial Value Problem Symbolically

开发者 https://www.devze.com 2023-03-25 11:42 出处:网络
I need to evaluate dsolve(\'Dy = 1 + exp(y)\', \'y(0) = 1\', \'t\') symbolically to conclude that t* >= log(exp(1) + 1) - 1.

I need to evaluate

dsolve('Dy = 1 + exp(y)', 'y(0) = 1', 't')

symbolically to conclude that

t* >= log(exp(1) + 1) - 1.

However, when I run it, I get a warning saying it can't开发者_运维百科 be solved explicitly and throws an [empty sym].

Similar equations don't seem to have this problem. Anyone have any ideas as to why it's not evaluating?


Here's the solution that Wolfram Alpha gives:

Use that to calibrate your expectations.


very unstable equation.... as an alternative, you can do:

e = exp(1);
yfun = dsolve('Dy = 1 + e^y', 'y(0) = 1', 't');
t = 0:0.01:0.3;
eval(yfun);
0

精彩评论

暂无评论...
验证码 换一张
取 消