I am attempting to solve the equation  $$f(x + 1) = f'(x)$$  for distributions $C \rightarrow C: f(x)$  My first guess to exploit the fact that this seems similar to identity  $$\sin\left( \frac{\pi}{2} - x\right) = \cos(x) = \frac{d}{dx} [\sin(x)] $$  I therefore assume my answer takes on the form:  $$Ce^{ax}$$   And now attempt to solve:  $$Ce^{a(x+1)} = Cae^{ax}$$  Yielding  $$e^{a}e^{ax} = ae^{ax}$$  Yielding  $$e^a = a$$  How do I extract a single solution, let alone the infinitely many complex solutions for $a$ that satisfy this equation?
12799
Let $f \colon \mathbb{R} \rightarrow  \mathbb{R}$ be a smooth function such that $$f'(x) = f(1+x)$$ How can we find the general form of $f$? I thought of some differential equations, but not sure how to use them here. Thanks.
42806
This seems like it should be super simple but I can't figure it out. Prove that if $f(t)$ satisfies $f(t)=f'(t)$, then $f(t+s)=f(t)f(s)$ for all $t,s$ belonging to R without using the fact that $f(t)=c\cdot e^t$. The only hint I'm given is that I can use the fact that two functions satisfying the same ODE $x'=x$ with the same initial condition are equal. Any ideas?
3511
$$e^x − 6e^{-x} − 1 = 0$$  No idea how to solve this. If someone could show me the first one or two steps to push me in the right direction that would be great.
81178
I'm trying to solve the integral of $\left(\int e^x\,\sin x\,\,dx\right)$ (My solution):  $\int e^x\sin\left(x\right)\,\,dx=$  $\int \sin\left(x\right) \,e^x\,\,dx=$  $\left(\sin(x)\,\int e^x\right)-\left(\int\sin^{'}(x)\,\left(\int e^x\right)\right)$  $\left(\sin(x)\,e^x\right)-\left(\int\cos(x)\,e^x\right)$  $\left(\sin(x)\,e^x\right)-\left(\cos(x)\,e^x-\left(\int-\sin\left(x\right)\,e^x\right)\right)$  $\left(\sin(x)\,e^x\right)-\left(\cos(x)\,e^x-\left(-\sin\left(x\right)\,e^x-\int-\cos\left(x\right)\,e^x\right)\right)$  I don't know how to complete because the solution gonna be very complicated.
13545
Problem:   If $f(x)$ is continous at $x=0$, and $\lim\limits_{x\to 0} \dfrac{f(ax)-f(x)}{x}=b$, $a, b$ are constants and $|a|&gt;1$, prove that $f'(0)$ exists and $f'(0)=\dfrac{b}{a-1}$.   This approach is definitely wrong:    \begin{align} b&amp;=\lim_{x\to 0} \frac{f(ax)-f(x)}{x}\\ &amp;=\lim_{x\to 0} \frac{f(ax)-f(0)-(f(x)-f(0))}{x}\\ &amp;=af'(0)-f'(0)\\ &amp;=(a-1)f'(0) \end{align}  I will show you a case why this approach is wrong:     \[f(x)= \begin{cases} 1,&amp;x\neq0\\ 0,&amp;x=0 \end{cases}\]   $\lim_{x\to0}\dfrac{f(3x)-f(x)}{x}=\lim_{x\to0} \dfrac{1-1}{x}=0$   but $\lim_{x\to0}\dfrac{f(3x)}{x}=\infty$,$\lim_{x\to0}\dfrac{f(x)}{x}=\infty$   Does anyone know how to prove it? Thanks in advance!
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Let , $y(x)$ be a continuous solution of the initial value problem $y'+2y=f(x)$  , $y(0)=0$ , where, $$f(x)=\begin{cases}1 &amp; \text{ if } 0\le x\le 1\\0 &amp; \text{ if } x&gt;1\end{cases}$$Then, $y(3/2)$ equals to (A) $\frac{\sinh(1)}{e^3}$ (B) $\frac{\cosh(1)}{e^3}$ (C) $\frac{\sinh(1)}{e^2}$ (D) $\frac{\cosh(1)}{e^2}$  Integrating both sides we get , $$\int_0^xd(y(x))+2\int_0^xy(x)\,dx=\int_0^xf(x)\,dx$$ $$\implies y(x)+2\int_0^xy(x)\,dx=\int_0^xf(x)\,dx=1$$ $$\implies y(x)=1-2C$$where , $$C=\int_0^xy(x)\,dx=(1-2C)x\implies C =\frac{x}{1+2x}$$ Then , $y(x)=\frac{1}{1+2x}$ and so, $y(3/2)=1/4$.  But no option match..Please help..
16540
I have a differential equation system  $x_1'(t) = -x_2(t)$  $x_2'(t) = -x_1(t)$  I see that I can write $\dot{x} = Ax$ where  $A = \begin{pmatrix}0 &amp; -1 \\ -1 &amp; 0\end{pmatrix}$  The complete real solution will be  $x(t) = c_1 e^{-t} \begin{pmatrix}1 \\ 1\end{pmatrix} + c_2 e^t \begin{pmatrix}-1 \\ 1\end{pmatrix}$  where $c_1,c_2 \in \mathbb{R}$.  It this correct?  Now I need to find the solution such that $x_1(0) = -1$ and $x_2(0) = 1$.  How can I do this?  I think I could split my complete solution in  $x_1(t) = c_1 e^t - c_2 e^{-t}$  and  $x_2(t) = c_1 e^t + c_2 e^{-t}$  and then find $c_1$ and $c_2$ by solving $x_1(0) = -1$ and $x_2(0) = 1$ with respect to $c_1$ and $c_2$ and finally substitute these values in the complete solution $x(t)$.  Is this correct?
84034
I need help solving this past exam question, the professor posted exam questions with out solutions to last year's exam, I tried reduction of order, however that seemed to complicate things. And then I tried Euler-Cauchy method however I realized that won't work due to the $(2x^2 + 1)y''$.  Given $y_1=x$ is a solution $$(2x^2 + 1)y'' − 4xy' + 4y = 0$$ Find the second solution $y_2$ for equation.
68251
How would I find the closed form of  $$ f(\alpha) = \sum_{n=0}^{\infty} \frac{\cos(\alpha \, n)}{n!}$$  My old pal Wolfram tells me that  $$ f(\alpha) = \mathrm{e}^{\cos(\alpha)}\cos(\sin(\alpha))$$  I've attempted writing out the first couple terms for different values of $\alpha$, but I haven't quite figured out how to arrive at the result.
109376
I recently came across this equation : $$\forall x \in \mathbb{R}_+^*, f'(x) = f\left(\frac1{x}\right)$$where $f \in \mathcal{C}^1(\mathbb{R}, \mathbb{R})$.  I've done the following, but I'm stuck at the end. Could you give me pointers? Thanks!  Differentiating yields $$\forall x, f''(x) = -\frac1{x^2}f(x) \tag{$S_0$}$$Solutions in the form $$x \mapsto \frac1{x^\phi}$$ work iff $\phi(\phi+1) = -1 $, ie. $\phi = \frac{-1 \pm i \sqrt{3}}{2} =e^{\pm 2i\pi/3} = j, \overline{j}$. Elements of the vector space generated by the free pair $(x^j, x^\overline{j})$ are therefore solutions of ($S_0$).  I then feed $\lambda x^j + \mu x^\overline{j}$ in the original equation, which yields $-\lambda j\frac1{x^{j+1}}-\mu\overline{j}\frac1{x^{\overline{j} + 1}} = \frac{x^{j + \overline{j}}}{\lambda x^\overline{j} + \mu x^j}$, then $(-\lambda j x^{\overline{j}+1} - \mu \overline{j} x^{j+1})(\lambda x^{\overline{j}} + \mu x^j) = x^{1+j+\overline{j}} = x^0 = 1$, and $-\lambda^2 j x^{2\overline{j} + 1} - \mu^2 \overline{j} x^{2j+1} - \lambda\mu(j + \overline{j}) = 0 $. Thus, $$ \lambda^2 j x^{-2i\sin(2\pi/3)} + \mu^2 \overline{j} x^{2i\sin(2\pi/3)} = \lambda\mu$$  Does that mean that no solutions can be found to the original equation, except the trivial $x \mapsto 0$ one? Or that I didn't take the right approach? I can't figure out how to handle the last equality.
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For which $a \in \mathbb{C} -\ \{0,1\}$ does the equation $f(z) = f(az) $ has a non constant solution $f$ with $f$ being analytical in a neighborhood of $0$.  My attempt:  First, we can see that any such solution must satisfy:  $f(z)=f(a^kz)$ for all $k \in \mathbb{N} $.  If $|a|&lt;1$:  The series $z_{k} = a^k$ converges to 0 which is an accumulation point, and $f(z_i)=f(z_j)$ for all $i, j\in \mathbb{N} $. Thus $f$ must be constant.  If $|a|=1$:  For all $a \neq 1$ , $f$ must be constant on any circle around $0$, so again $f$ must be constant.  My quesions are:   Am I correct with my consclusions?   Also, I'm stuck in the case where $|a|&gt;1$. Any ideas?  Thanks
88391
The goal is to show (preferably without contour integration, as my knowledge is pretty limited there, but if you can do it that way there is no problem to share it) that: $$I= \int_0^1 \frac{x\ln\left(\frac{1+x}{1-x}\right)}{\left(\pi^2+\ln^2\left(\frac{1+x}{1-x}\right)\right)^2}\mathrm dx=\frac{1}{240}$$  I am still trying to find more relevant approaches, but so far I thought of: Considering $$J(a)= \int_0^1 \frac{x\ln\left(\frac{1+x}{1-x}\right)}{a^2+\ln^2\left(\frac{1+x}{1-x}\right)}dx$$ We have that $J'(\pi)=-2\pi I$, also $$J(a)=\int_0^\infty e^{-at}\int_0^1 x\sin\left(t\ln\left(\frac{1+x}{1-x}\right)\right)dx dt $$ Which is nice, however I couldnt solve the inner integral.  Another approach would be to let $\ln\left(\frac{1+x}{1-x}\right)=t \rightarrow x=\frac{e^t-1}{e^t+1}\rightarrow dx=\frac{2e^t}{(e^t+1)^2}$  $$I=2\int_0^\infty \frac{e^t(e^t-1)}{(e^t+1)^3}\frac{t}{(\pi^2+t^2)^2}dt$$ This doesnt look nice, but notice that $$\frac{d^2}{dx^2} \left(\frac{1}{1+e^x}\right)=\frac{e^x(e^x-1)}{(e^x+1)^3}$$ Integrating by parts two times I get $$I=\frac{1}{\pi^2} +24 \int_0^\infty \frac{x^2-\pi^2}{(x^2+\pi^2)^4}\frac{x}{1+e^x}dx$$ Now I am thinking to use series for $\frac{x}{1+e^x}$, but I dont know a series that satisfies it, similary we have for $\frac{x}{e^x-1}$ in terms of Bernoulli numbers. Could you land some help for this integral?
8683
$x′ = Ax$  where $$A = \begin{pmatrix} 3&amp;1&amp;0\\ 0&amp;3&amp;0\\ 0&amp;0&amp;2 \end{pmatrix} $$ what is the logic to solve such type of equations?  I found that $det(A-λI) = (3-λ)^2(2-λ)=0$, so the eigenvalues are 3 and 2
64922
We already solved or argued differential equations, integral equations and even function equations. But I encounter the following, which I call it "series equation":     Is there any nonzero function $f$ satisfying the series equation   $$f(x)=\sum_{n=0}^{\infty}\frac{f(n)}{n!}x^n \tag{*} $$   The source: I interested to solve the equation $$f'(x)=f(x+1)$$ and I found out $$f^{(n)}(x)=f(x+n)$$ thus: $f^{(n)}(0)=f(n)$ and if we consider  $f$ having a taylor series, then the expression $(*)$ will be the case.  But I have no idea how $f$ would be. Is such a function exists?  Thanks.
4007
$f(x) = \frac{x}{x^2 + 1}$.  How do you find $f^{-1}$?  Is try and solve for y and I get the following:  $y = \frac{x}{x^2 + 1}$  $y(x^2 + 1) = x$  $yx^2 + y = x$  $y = -yx^2 + x$  $y = x(-yx + 1)$  It seems whatever I do I can't isolate the x variable, I just go in circles...
80626
I don't know how to attack this problem. The last I've tried is using a differential equation, but I don't know how to solve it.  Let $y$ be $f^{-1}(x)$. Knowing that $x=f(y)= 2y + \cos(y)$ and derivating I obtained the following non-linear first order differential equation: $y' \cdot (2-\sin(y))=1$  I would thank you if you can help me.  Edit: I haven't said, but it is trivial to check that the function is injective, so it has an inverse, because $\forall x \in \mathbb{R}$ $f'(x) \neq0$
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When solving the differential equation $y'' + ay' + by = 0$ (with constant, real coefficients $a$ and $b$, although they could be complex if you like), you do it by setting up the characteristic equation $r^2 + ar + b = 0$, finding its solutions $r_1, r_2$, and then the general solution to this equation is $Ce^{r_1x} + De^{r_2x}$. This works both when the solutions are real and when they are complex.  However, when we have a double root $r_1 = r_2$, we get a different general solution, namely $Ce^{r_1x} + Dxe^{r_1x}$. I have no trouble seing that this is indeed a solution, and intuitive reasoning on degrees of freedom dictates that we must have a linear combination of two terms in our general solution, while $e^{r_1x}$ and $e^{r_2x}$ are the same. So the fact that there is a second term of some other form is not surprising.  I have, however, yet to see a "natural" explanation of this $xe^{r_1x}$ term. If one were developing the theory from scratch, how would one find this solution (other than blind luck)? If I wanted to teach ODE's to a class of students "the right way", i.e. with good explanations and motivations for everything (as opposed to just pulling out ready-made solutions like what was done to me when I was learning this exact thing), how would I motivate even considering a term like $xe^{r_1x}$ (other than "Well, exponentials aren't quite cutting it, but this is kindof like an exponential, right? Let's try it.")? And is there a way of solving the general differential equation that does not involve splitting into cases depending on whether the characteristic polynomial is a square?