Hi. Solving $3y'=x+2xy$, I got $\ln \left | 1+2y \right |=\frac{x^{2}}{3}+C$ after integrating, and finally arrived at
$$y=\pm \frac{e^{\frac{x^{2}}{3}+C}-1}{2}=\pm\frac{e^{\frac{x^{2}}{3}+C}-1}{2}\mp \frac{1}{2}$$.
I checked my result using WolframAlpha, and the solution it came up with was
$$y=Ce^{\frac{x^{2}}{3}}-\frac{1}{2}$$.
That confuses me. I know that we can eliminate $\frac{e^{C}}{2}$ by choosing a constant $D$ such that $D=\frac{e^{C}}{2}$, and since $D$ can be both positive or negative, the $\pm$ in front of $\frac{e^{\frac{x^{2}}{3}+C}-1}{2}$ can be eliminated as well, but how…
$$y=\pm \frac{e^{\frac{x^{2}}{3}+C}-1}{2}=\pm\frac{e^{\frac{x^{2}}{3}+C}-1}{2}\mp \frac{1}{2}$$.
I checked my result using WolframAlpha, and the solution it came up with was
$$y=Ce^{\frac{x^{2}}{3}}-\frac{1}{2}$$.
That confuses me. I know that we can eliminate $\frac{e^{C}}{2}$ by choosing a constant $D$ such that $D=\frac{e^{C}}{2}$, and since $D$ can be both positive or negative, the $\pm$ in front of $\frac{e^{\frac{x^{2}}{3}+C}-1}{2}$ can be eliminated as well, but how…
By Problem 6-3, there is a smooth function $\tilde{\delta}:M\to\Bbb R$ such that $\tilde{\delta}|_B\equiv 0$ and $0<\tilde{\delta}(x)<\delta(x)$ on $M\setminus B$. Since $M\setminus B$ is an embedded submanifold of $M$, by smooth approximation theroem on $f|_{M\setminus B}$, there is a smooth map...
I'm trying to prove Lemma $2$ in this paper. The statement is as follows:
Let $\{p_j\}$ be a sequence (possibly finite) of non-negative numbers, and $\{-q_j\}$ be an infinite sequence of negative numbers such that $\sum p_j = \sum q_j \leqslant \pm\infty$, $q_j \to 0$ as $j\to\infty$, and $p_j \...
How can I prove that for a natural number $k$ we have the following?
$$(-1)^k\binom{1/2}k=-\frac1{(2k-1)2^{2k}}\binom{2k}k$$
It seems like we use a rule for re-writing which I cannot see.
