Prove that the ordinary differential equation $$y'=-y^3+\sin x $$ has a unique solution satisfying $y(0)=y(2\pi)$ It's really hard for me to figure out the existence and the uniqueness. I don't have any clue about this problem. Maybe we can let $u(x)=y(x+2\pi)-y(x)$, then $u(x)$ should satisfy $u(0)=0$ and some differential equations.
124029
So, for the ODE \begin{align} x' = -x^{3} + \sin t, \end{align} we can show that there exists a $2\pi$ periodic solution.  To do this, we denote by \begin{align} x(t,\alpha) \end{align} The solution $x(t)$ of the ODE such that $x(0) = \alpha$.  Then, let $\alpha \in [-2,2]$.  Consider the function $f(\alpha) = x(2\pi, \alpha)$ \begin{align} x &gt; 1 &amp;\Rightarrow x' &lt; 0 \\ x &lt; 1 &amp;\Rightarrow x' &gt; 0. \end{align}  Thus, solutions beginning in $[-2,2]$ stay there and we can use Brouwer's fixed point theorem to show that there exists a fixed point of $f$.  Therefore there is some $\alpha^{*}$ such that $f(\alpha^{*}) = \alpha^{*}$, which represents the periodic solution $x(t,\alpha)$.  Now, my question: Is this $2\pi$- periodic solution unique? Are there other $2\pi$ periodic solutions of this ODE?  I have started by trying to subtract two periodic solutions from one another but this hasn't taken me anywhere productive.
108117
Find all functions $g:\mathbb{R}\to\mathbb{R}$ with $g(x+y)+g(x)g(y)=g(xy)+g(x)+g(y)$ for all $x,y$.  I think the solutions are $0, 2, x$. If $g(x)$ is not identically $2$, then $g(0)=0$. I'm trying to show if $g$ is not constant, then $g(1)=1$. I have $g(x+1)=(2-g(1))g(x)+g(1)$. So if $g(1)=1$, we can show inductively that $g(n)=n$ for integer $n$. Maybe then extend to rationals and reals.
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I'm having this question in my homework assignment in Linear Algebra and diffrential equation class, and trying to find the general solution for this second ODE.  $$y''y^3 = 1$$  Using substitution I said $p = y'$ and $p' = y''  \rightarrow  \frac{dp}{dx}= \frac{dp}{dy} \times \frac{dy}{dx}$  Then we get $y^3 \frac{dp}{dx}$ now what's the next step should I integrate or differentiate the equation? What's the trick on these type of ODEs?  Thanks in advance!
107151
Show that there exists a unique $v_0 \in H^1(0,1)$ such that $u(0)=\int_0^1(u'v_0'+uv_0), \forall u \in H^1(0,1)$. Further Show that $v_0$ is the solution of some differential equation with appropriate boundary conditions. Compute $v_0$ explicitly.   Let $f: H^1(0,1) \to \mathbb{R}$ be defined by $f(u)=u(0)$. Then I showed that $f$ is linear and continuous. Hence there exists $v_0 \in H^1(0,1)$ such that $u(0)=\int_{0}^1u'v_0'+\int_{0}^1uv_0$  The differential equation is given by $f=-u''+u$ with the conditions $u'(0)=u'(1)=0$.   How do I compute $v_0$ explicitly??
120128
I have shown that a smooth solution of the problem $u_t+uu_x=0$ with $u(x,0)=\cos{(\pi x)}$ must satisfy the equation $u=\cos{[\pi (x-ut)]}$. Now I want to show that $u$ ceases to exist (as a single-valued continuous function) when $t=\frac{1}{\pi}$.  When $t=\frac{1}{\pi}$, then we get that $u=\cos{(\pi x-u)}$.  With single-valued function is it meant that the function is 1-1 ?  If so, then we have that $\cos{(2 \pi-u)}=\cos{(4 \pi -u)}$, i.e. for two different values of $x$, we get the same $u$, and so for $t=\frac{1}{\pi}$, $u$ is not 1-1.  But if this is meant, how are we sure that for $t \neq \frac{1}{\pi}$ the function is single-valued?
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Here is the ODE equation $xy' = 2y$. A book states that the interval where it makes sense doesn't include $0$. Can you explain why? As to me, I think it is defined everywhere.  This problem is from the book "Ordinary Differential Equations - Harry Pollard, Morris Tenenbaum". The exact problem states: "Prove that the functions in the right-hand column below are solutions of the differential equations in the left-hand columns. (Be sure to state the common interval for which solution and differential equation make sense.)"  $xy' = 2y$ and $y = x^2$ with the answer $x \neq 0$
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Solve the differential equation: $yy'^2-2xy'+y=0$ (*)  I have   $(*)\Leftrightarrow (yy')^2-2xyy'+y^2=0\\ \Leftrightarrow z'^2-4xz+4z^2 =0 \ \ (z=y^2)$  I can't know how to continue.  Could you give me some hints. Thanks for helping.
102569
Find the solution of following differential equation:     $$\frac{dy}{dx}=\frac{1}{xy(x^2 \sin (y^2)+1)}$$   Could someone hint me something to get through this problem?
78760
Express the 2nd order ODE $$\begin{align}\mathrm d_t^2 u:=\frac{\mathrm d^2 u}{\mathrm dt^2}&amp;=\sin(u)+\cos(\omega t)\qquad \omega \in \mathbb Z /\{0\} \\u(0)&amp;=a\\\mathrm d_t u(0)&amp;=b\end{align}$$ as a system of 1st order ODEs and verify there exists a global solution by invoking the global existence and uniqueness theorems.  I'm not sure how to express second order ODEs as first order ODEs, any tips?
120773
[1] $y'' - 4y' + 4y = 0$  Usually problem like these will have the answer in the form $C_1e^a + C_2e^b ... $ where $a $ and $b$ are the roots of the characteristic equation $e^{rt}$  $$ y = e^{rt} $$ $$ y' = re^{rt}.. y'' = r^2 e^{rt} $$  $$ r^2e^{rt} - 4e^{rt} + 4e^{rt} = 0$$ $$ e^{rt}(r-2)(r-2) = 0$$ $$ y= C_1e^{2t} + C_2e^{2t}$$  However, this is not correct as the answer is $ y = C_1e^{2t} + C_2te^{2t}$ ! I just don't know why.  [2] For a similar problem , $y'' + 3y' - 4y = 0 $ I did the exact same thing and the answer is $$y = c_1e^t + c_2e^{-4t} $$  How are [1] and [2] different? They look the same, why does [2]'s solution have an extra factor of t.
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Specifically, if we know that one solution is $y_1(t)$, then why is the second solution in the form $y_2(t) = v(t) y_1(t)$?  Where $v(t)$ is the function that you need to solve for.  Why does this assumption always work?
2958
Why is $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha$? How can we find the RHS if we don't know what it is? (instead of proving the identity itself) I could find a geometric solution in , but is there any solution that doesnt require drawing something? Edit: I now saw some nice proofs using $e$ and Euler's identity. I would appreciate anything NOT using them too, as a change. Out of curiosity, Is there a proof that says you actually "can't" prove the identity using only other simple 1-variable identities like $\sin^2\alpha+\cos^2\alpha=1$?
74360
Show that the solution for the Diophantine equation $x^2 - y^2 = N$ is unique if and only if $\mid N \mid$ or $\frac{\mid N \mid}{4}$, respectively, is $1$ or prime.   I have an idea of how to show $\mid N \mid$ or $\frac{\mid N \mid}{4}$ is $1$ or prime $\Rightarrow$ the solution for the Diophantine equation is unique.  We can show this by going through the separate cases $ \mid N \mid = 1$ or $p$ and $\frac{\mid N \mid}{4} = 1 $ or $p$. For example, if $\mid N \mid = 1$, then $x^2 - y^2 = (x -y)(x+y) = 1$ or $(x -y)(x+y) = -1 $. For $(x -y)(x+y) = 1$, we have $x = 1, y = 0$ and $x = -1, y = 0$ as solutions (more than one solution? Doesn't uniqueness imply only one?).  How do I show the other direction?: The solution for the Diophantine equation is unique $\Rightarrow$ $\mid N \mid$ or $\frac{\mid N \mid}{4}$ is $1$ or prime.
80845
Are exponential &amp; trigonometric functions the only non-trivial solutions to $F'(x)=F(x+a)$?   $F(x)=0$ would be the trivial solution. Then, for $a=0$ (or  $a=2\pi i$), we have $F(x)=e^x$, and for $a=\dfrac\pi2$ there are $F(x)=\sin x$ and $F(x)=\cos x$. But the three are connected by  $e^{ix}=\cos x$ $+i\sin x$. Indeed, on a more  note, letting $F(x)=e^{\lambda x}$, we have $\lambda=\dfrac{W(-a)}{-a}$ where W is the . My question would be if these are the only ones, due to the special properties of the number e and the exponential function, or if there aren't by any chance more, which do not belong in the same family or category as these, i.e., which are not exponential or trigonometric in nature ? Thank you.
91334
Show that $\sin(z) = z$ has infinitely many solutions in complex numbers. Little Picard theorem should help, but using big Picard theorem is undesirable. Thanks a lot!
75098
I understand how to find the general solution of a differential equation, for complex roots, simple roots, nonhomogeneous equations etc... but I still don't really know what the general solution is?   As in, what is the purpose of it? What is it used for? thanks...
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.