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June 17th, 2018, 12:39
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  106 view(s)

June 17th, 2018, 00:59
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  106 view(s)

June 17th, 2018, 00:52
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  106 view(s)

June 16th, 2018, 19:51
\ln(\cos(x))=\ln\left((\cos(x))^{1}\right)=\ln(\sec(x))
8 replies  106 view(s)

June 16th, 2018, 19:38
\mu(x)=\exp\left(\int \tan(x)\,dx\right)=e^{\Large\ln(\sec(x))}=\sec(x)
8 replies  106 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...
4 replies  98 view(s)

June 15th, 2018, 16:28
\mu(x)=\exp\left(\int\frac{1}{x}\,dx\right)=e^{\ln(x)}=x
3 replies  59 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  59 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  69 view(s)

June 15th, 2018, 12:44
We could also simply state:
\d{x}{y}=e^yx
\d{x}{y}+x=e^y
Now, our integrating factor is:
\mu(y)=\exp\left(\int\,dy\right)=e^y
4 replies  69 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}{ux}
Or:
4 replies  69 view(s)

June 14th, 2018, 11:23
Hello and welcome to MHB! (Wave)
I've moved this thread to our elementary algebra forum, since this is a better fit for the problem and I've give...
2 replies  74 view(s)

June 11th, 2018, 21:08
You now have the correct integrating factor, but you didn't multiply the RHS by this factor. :)
You should eventually get:
...
4 replies  74 view(s)

June 11th, 2018, 19:17
Your integrating factor is incorrect. Try finding that again. :)
4 replies  74 view(s)

June 11th, 2018, 15:20
The integrand:
\frac{e^x}{x^2+1}
does not have an antiderivative in elementary terms, and so the solution is given as a definite integral. The...
1 replies  47 view(s)

June 10th, 2018, 02:32
I think what you mean is:
\frac{d}{dx}\left(e^{2x}(2y)\right)
Please perform the indicated differentiation (product rule), and see if you get...
5 replies  118 view(s)

June 10th, 2018, 01:52
Your integrating factor is correct, but I would use the notation:
\mu(x)=\exp\left(2\int \,dx\right)=e^{2x}
5 replies  118 view(s)

June 10th, 2018, 00:24
Please check your computation...I get a different result. :)
7 replies  99 view(s)

June 9th, 2018, 22:55
I would write the general solution to the ODE as:
y(x)=2(x1)e^{2x}+c_1e^x
Now, we can write:
y(0)=2(01)e^{2\cdot0}+c_1e^0=1
Solve this...
7 replies  99 view(s)
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