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Today, 00:02

MarkFL replied to a thread 2.2.8 de xy'+2y=sin x,y(pi)=1/pi find solution in Differential Equations

You wound up with $x^2\sin(x)$ on the RHS before integrating, but it should be $x\sin(x)$. :)

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2 replies | 54 view(s)
Yesterday, 19:23

MarkFL replied to a thread [SOLVED] 2.2.7 de with trig and y(1)=1 in Differential Equations

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 ...

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2 replies | 36 view(s)
June 19th, 2018, 19:21

MarkFL replied to a thread 2.2.6 de with e^x in Differential Equations

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 $e^x$.

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1 replies | 41 view(s)
June 19th, 2018, 15:31

MarkFL replied to a thread 2.2.5 de equation with y(1)=1/2 in Differential Equations

Yes, but you used it on the original ODE, not the one in standard linear form. :)

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4 replies | 91 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.

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8 replies | 113 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...

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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: ...

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8 replies | 113 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))

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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)

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8 replies | 113 view(s)
June 16th, 2018, 00:29

MarkFL replied to a thread Problems with Percentage in Pre-Algebra and Algebra

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...

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6 replies | 126 view(s)
June 15th, 2018, 16:28

MarkFL replied to a thread [SOLVED] 2.2.1 de with u substitution in Differential Equations

\mu(x)=\exp\left(\int\frac{1}{x}\,dx\right)=e^{\ln(x)}=x

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3 replies | 63 view(s)
June 15th, 2018, 15:53

MarkFL replied to a thread [SOLVED] 2.2.1 de with u substitution in Differential Equations

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: ...

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3 replies | 63 view(s)
June 15th, 2018, 14:14

MarkFL replied to a thread 2.1.9 de with reciprocal in Differential Equations

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...

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4 replies | 73 view(s)
June 15th, 2018, 12:44

MarkFL replied to a thread 2.1.9 de with reciprocal in Differential Equations

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

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4 replies | 73 view(s)
June 15th, 2018, 03:08

MarkFL replied to a thread 2.1.9 de with reciprocal in Differential Equations

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:

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4 replies | 73 view(s)
June 14th, 2018, 11:23

MarkFL replied to a thread Rationalize expression in Pre-Algebra and Algebra

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...

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2 replies | 89 view(s)
June 11th, 2018, 21:08

MarkFL replied to a thread 2.1.{8} de with trig y(pi)=0 in Differential Equations

You now have the correct integrating factor, but you didn't multiply the RHS by this factor. :) You should eventually get: ...

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4 replies | 78 view(s)
June 11th, 2018, 19:17

MarkFL replied to a thread 2.1.{8} de with trig y(pi)=0 in Differential Equations

Your integrating factor is incorrect. Try finding that again. :)

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4 replies | 78 view(s)
June 11th, 2018, 15:20

MarkFL replied to a thread 2.1.7 Find the solution of the given initial value problem. in Differential Equations

The integrand: \frac{e^x}{x^2+1} does not have an anti-derivative in elementary terms, and so the solution is given as a definite integral. The...

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1 replies | 50 view(s)
June 10th, 2018, 02:32

MarkFL replied to a thread [SOLVED] 2.1.6 de with IVP in Differential Equations

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...

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5 replies | 121 view(s)
June 10th, 2018, 01:52

MarkFL replied to a thread [SOLVED] 2.1.6 de with IVP in Differential Equations

Your integrating factor is correct, but I would use the notation: \mu(x)=\exp\left(2\int \,dx\right)=e^{2x}

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5 replies | 121 view(s)
June 10th, 2018, 00:24

MarkFL replied to a thread 2.1.5 de with e in Differential Equations

Please check your computation...I get a different result. :)

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7 replies | 101 view(s)
June 9th, 2018, 22:55

MarkFL replied to a thread 2.1.5 de with e in Differential Equations

I would write the general solution to the ODE as: y(x)=2(x-1)e^{2x}+c_1e^x Now, we can write: y(0)=2(0-1)e^{2\cdot0}+c_1e^0=1 Solve this...

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7 replies | 101 view(s)

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We are an online community that gives free mathematics help any time of the day about any problem, no matter what the level. You will have to register before you can post. To start viewing messages, select the forum that you want to visit from the selection below.

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June 21st, 2018

June 20th, 2018

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MarkFL

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