how to write in matlab
y = e^ int(75/80)dt

5 个评论

Please provide more detail. What differential equation are you trying to solve?
d dtx1(t) = − x1(t) 2 + 20 d
solved using linear DE equation to get
y = exp^int(1/2)dt ** not 75/80
James Tursa
James Tursa 2018-12-12
编辑:James Tursa 2018-12-12
Is it this? (where x1 is a function of time)
d(x1)/dt = -x1 / 2 + 20 * d
What is that d at the end?
sorry no d so just
d(x1)/dt = -x1 / 2 + 20
Stephan
Stephan 2018-12-12
编辑:Stephan 2018-12-12
Hi Brenda,
i saw that you got some answers on your severa questions here in the forum. Did you notice that you can accept and/or vote for answers, in order to thank the volunteers here for taking the time to help you with your problems?

请先登录,再进行评论。

 采纳的回答

madhan ravi
madhan ravi 2018-12-12
编辑:madhan ravi 2018-12-12
You can verify the result with wolfram but here you will see a C which is constant which arises after antiderivative(without limits):
syms y(t)
ode=diff(y) == exp(75/80)
dsolve(ode)

9 个评论

what about the exponential before the integral how to represent that
exp() see edited answer
This is how you write the equation:
syms t y
eqn = y == vpa(exp(1)^(int(75/80,t))) % in decimal form
eqn = y == exp(1)^(int(75/80,t)) % in fractions form
Brenda Galabe's answer moved here for consistency:
but the equation looks like this
y = exp^int(75/80)dt . so the exp is before the integral
Brenda Galabe's answer moved here for consistency:
ok cool. answer im getting is
eqn =
y == (3060513257434037/1125899906842624)^((15*t)/16)
i know on paper answer suppose to be exp^(75/80)t
Basically they are the same but represented in different forms just substitute the values and see as below:
>> t=1;
>> exp(1)^(75/80)*t
ans =
2.5536
>> (3060513257434037/1125899906842624)^((15*t)/16)
ans =
2.5536
>>
Brenda Galabe's answer moved here for consistency:
ok thank you
Anytime :)

请先登录,再进行评论。

更多回答(1 个)

Brenda Galabe
Brenda Galabe 2018-12-12

0 个投票

not sure what that means

类别

帮助中心File Exchange 中查找有关 Mathematics 的更多信息

产品

版本

R2015b

标签

Community Treasure Hunt

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

Start Hunting!

Translated by