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• Today, 00:02
You wound up with $x^2\sin(x)$ on the RHS before integrating, but it should be $x\sin(x)$. :)
2 replies | 54 view(s)
• Yesterday, 19:23
When you multiply through by $\mu(x)$, you should have: \frac{d}{dx}(\sin(x)y)=2 And then integrate: \sin(x)y=2x+c_1 ...
2 replies | 36 view(s)
• June 19th, 2018, 19:21
Towards the end, when you divide through by $x$, you want: y(x)=\frac{e^x}{x}+\frac{c}{x} You mistakenly divided the constant by $e^x$.
1 replies | 41 view(s)
• June 19th, 2018, 15:31
Yes, but you used it on the original ODE, not the one in standard linear form. :)
4 replies | 91 view(s)
• June 18th, 2018, 06:19
Thanks steenis ... No worries at all ... Thanks for all your help ... Peter
10 replies | 220 view(s)
• June 18th, 2018, 06:04
Thanks Steenis ... That proof seems really clear ... Will work through it again shortly... Peter
4 replies | 82 view(s)
• June 18th, 2018, 05:18
Sorry Steenis ... I don't understand you ... Can you give me a hint as to what is wrong ...? Peter
10 replies | 220 view(s)
• June 18th, 2018, 04:51
Thanks steenis ... most helpful ... Can see that the short exact sequence $0\rightarrow \text{ker } f \overset{i}{ \rightarrow}R^{(n)}... 10 replies | 220 view(s) • June 18th, 2018, 01:22 ======================================================================== Since I could not see any specific errors, I have completed the proof... 4 replies | 82 view(s) • June 17th, 2018, 12:39 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations I can't imagine trying to use the internet on a telephone. I'm sorry scrolling on a telephone is such a chore...they should fix that. 8 replies | 113 view(s) • June 17th, 2018, 05:04 Peter started a thread Deveno ... in Chat Room Deveno is much missed ... especially by those who frequent the Linear and Abstract Algebra Forum ... Deveno's pedagogical abilities were as... 0 replies | 58 view(s) • June 17th, 2018, 04:40 I am reading Dummit and Foote's book: "Abstract Algebra" (Third Edition) ... I am currently studying Chapter 10: Introduction to Module Theory ...... 4 replies | 82 view(s) • June 17th, 2018, 00:59 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations Are you using Tapatalk by any chance? I have code in place to let me know when posts have been edited, so I don't miss added content (it's better to... 8 replies | 113 view(s) • June 17th, 2018, 00:52 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations When you multiply by$\mu(x)$, you get: \sec(x)y'+\tan(x)\sex(x)y=\sec(x)\sin(2x) This can be written as: ... 8 replies | 113 view(s) • June 16th, 2018, 23:10 Thanks Steenis ... You have shown that$R^{(n)} / N \cong M$where$N = \text{ Ker } f$... ... ... ... ... (1) ... and we have by... 10 replies | 220 view(s) • June 16th, 2018, 19:51 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations -\ln(\cos(x))=\ln\left((\cos(x))^{-1}\right)=\ln(\sec(x)) 8 replies | 113 view(s) • June 16th, 2018, 19:38 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations \mu(x)=\exp\left(\int \tan(x)\,dx\right)=e^{\Large\ln(\sec(x))}=\sec(x) 8 replies | 113 view(s) • June 16th, 2018, 00:29 300+300\cdot\frac{1666}{100}=300(1+16.66)=300\cdot17.66=5298 Here, we have taken 300, and added 1666% of 300 to it. However if we multiply 300 by... 6 replies | 126 view(s) • June 16th, 2018, 00:03 Thanks steenis ... but not sure if I follow .. ... but will try ... as follows ... We have an epimorphism$f:R^{(n)} \longrightarrow M$... 10 replies | 220 view(s) • June 15th, 2018, 16:28 \mu(x)=\exp\left(\int\frac{1}{x}\,dx\right)=e^{\ln(x)}=x 3 replies | 63 view(s) • June 15th, 2018, 15:53 We can see the integrating factor is$\mu(x)=x$and so the ODE will become: \frac{d}{dx}(xy)=x\sin(x) Upon integrating, we get: ... 3 replies | 63 view(s) • June 15th, 2018, 14:14 That's fine giving$x$as a function of$y$...I just chose to give$y$as a function of$x$since the original equation has$x\$ as the independent...
4 replies | 73 view(s)
• June 15th, 2018, 12:44
We could also simply state: \d{x}{y}=e^y-x \d{x}{y}+x=e^y Now, our integrating factor is: \mu(y)=\exp\left(\int\,dy\right)=e^y
4 replies | 73 view(s)
• June 15th, 2018, 03:08
I would begin here with the substitution: u=e^y\implies u'=uy' And so the ODE becomes: \frac{1}{u}u'=\frac{1}{u-x} Or:
4 replies | 73 view(s)
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