All Questions
15
questions
0
votes
3
answers
187
views
Solutions for $f'=\lambda f$
I am trying to figure out the following problem: Show that $f'=\lambda f$ for a real constant $\lambda$ has only $ce^{\lambda x}$ solutions.
My work: We take a look at $g(x)=f(x)\exp(\lambda x)$. We ...
- 81
3
votes
1
answer
309
views
Recommendations about the exponential function
I am studying differential equations and I am very surprised by how omnipresent the exponential function is. It pops up everywhere, but there isn't usually a lot of detail provided in introductory ...
- 167
0
votes
2
answers
71
views
Alternative argument to show that function diverges everywhere
Consider the function:
$$f(x) = x^{\frac{1}{2}} + \frac{1}{2}x^{-\frac{1}{2}} +\frac{2}{3}x^{\frac{3}{2}} - \frac{1}{4}x^{-\frac{3}{2}} +\frac{1}{15}x^{\frac{5}{2}} + \cdots$$
which is constructed ...
- 2,466
0
votes
1
answer
83
views
differential equations, exponential population growth
If p is population and t is time. Does that mean that when you do dp/dt = 0 you can find the maximum and minimum population
- 21
0
votes
1
answer
101
views
Can we express all functions in the exponential family with this differential equation?
Using the "prime" notation for differentiation $$f'(x) = \frac{\partial }{\partial x}\{f\}(x)\\f''(x) = \frac{\partial^2 }{\partial x^2}\{f\}(x)\\\vdots\\f^{(k)}(x) = \frac{\partial^k }{\partial x^k}\{...
- 26.1k
3
votes
7
answers
615
views
Intuitive explanation of $y' = y \implies y = Ce^x$
I understand why $f : \mathbb{R} \to \mathbb{R}$ with $f'(x) = f(x)$ and $f(0) = 1$ must be $f (x) = e^x$, but I don't really feel it is super intuitive. Intuitively, why would you expect such a ...
- 1,872
0
votes
2
answers
866
views
$f'(x)=af(x) \Rightarrow f(x)=e^{ax} f(0)$
I've been working on the following exercise: Let $f: \mathbb {R} \longrightarrow \mathbb{R}$ be differentiable. Suppose, there is an $a \in \mathbb{R}$ such that $$f'(x)=a \cdot f(x) ~~~~(*)$$
for all ...
- 1,003
0
votes
1
answer
66
views
ODE: $y'=3e^{2(x+y)}-1$, $y(0)=7$
How can I solve the following differential equation? $$y'=3e^{2(x+y)}-1,\quad y(0)=7$$
I am failing to separate the variables, and I am not yet introduced to other solving-methods.
Thanks in advance....
- 1,220
2
votes
2
answers
161
views
Finding sum of infinite series $1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\ldots $
So the question is 'express the power series $$1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\ldots $$
in closed form'.
Now we are allowed to assume the power series of $e^x$ and also we derived the ...
- 1,828
0
votes
1
answer
40
views
Starting from the assumption that $e^x$ is the solution to the equations $f(x)=f'(x)$ and $f(0)=1$, how may one derive the direct expansions of $e$?
If it is assumed that $e^x$ is the unique solution to the differential equation $f(x)=f'(x)$, how may we derive from the initial assumption the infinite sum expansion and the infinite product ...
- 981
2
votes
1
answer
100
views
A function and its derivative chasing tails
For which $t\ge0$ does there exist a differentiable function $f$ with $f(0)=0$, $f'(x)>f(x)$ for all $x>0$ and with $f'(0)=t$?
This question was inspired by (and is a variation of) the ...
- 13.5k
1
vote
3
answers
741
views
function bounded by an exponential has a bounded derivative?
here's the question. I want to be sure of that. Let $v:[0,\infty) \rightarrow \mathbb{R}_+$ a positive function satisfying
$$\forall t \ge 0,\qquad v(t)\le kv(0) e^{-c t}$$
for some positive constants ...
- 346
2
votes
3
answers
505
views
Find a particular solution of $\,\,y''+3y'+2y=\exp(\mathrm{e}^x)$
I already solved for the homogeneous one, but I'm still looking for the particular solution of the differential equation:
$$y''+3y'+2y=\exp(\mathrm{e}^x)$$
The homogeneous solutions of this system ...
- 984
0
votes
2
answers
50
views
If $z'\le az+b$ then $z(t)\le z_0+bt$
If $z$ satisfies; $z'\le az+b$, $\ z(0)=z_0>0$ with constants $a,b$ why is true that $z(t)\le z_0+bt$, if $a=0$
It is clear that it can't be justified only by integrating. We had only Gronwall ...
- 1,367
4
votes
3
answers
1k
views
Prove that if $\phi'(x) = \phi(x)$ and $\phi(0)=0$, then $\phi(x)\equiv 0$. Use this to prove the identity $e^{a+b} = e^a e^b$.
I am given the following. hint Consider $f(x)=e^{-x} \phi(x)$.
I am unsure how to approach this problem.
- 2,649