Author Topic: Hyperblade01 Question Thread (Read 6872 times) Tweet Share

hyperblade01 · « **on:** July 17, 2009, 09:05:07 pm »

To start off...

Solve the differential equation:

$\frac{dy}{dx} = 9-y^2$ given x = 0 and y = 2

$x = \int \frac{1}{9-y^2} dy$

$x = \frac{1}{6}log_e|\frac{y+3}{y-3}|+c$

$c = \frac{-1}{6}log_e5$

$x = \frac{1}{6}log_e|\frac{y+3}{5y-15}|$

$e^{6x} = \frac {y+3}{5y-15}$

$y = \frac {3+15e^{6x}}{5e^{6x}-1}$

Need confirmation if that answer is right (skipped alot of steps as cbs at the moment) and if the answer is wrong, just a note on whereabouts I went wrong

Thanks

dummy · « **Reply #1 on:** July 17, 2009, 09:27:16 pm »

$9-x^2=(3-x)(3+x)$

that's where i work differently. u might be right

TrueTears · « **Reply #2 on:** July 17, 2009, 09:30:24 pm »

Yeah that is fine hyperblade.

EDIT: actually this reminds me of /0 question about the mod.

ryley · « **Reply #3 on:** July 17, 2009, 09:35:37 pm »

Are you sure TrueTears?, looking at it quickly I'm not sure hyperblade's final answer satisfies the given condition, so from that I think he may have made a mistake.

TrueTears · « **Reply #4 on:** July 17, 2009, 09:40:06 pm »

Hmm, when I integrate $\frac{1}{9-y^2}$ I get the same result as he does.

Oh wait mistake is here, $\frac{y+3}{5y-15} = \pm e^{6x}$ because of the mod

So, $\pm e^{6x}$ $\implies$ $\frac{y+3}{5y-15} = -1$ when y = 2.

Therefore $-e^{6x} = -1$ since $e^{6x} \neq -1$

hyperblade01 · « **Reply #5 on:** July 17, 2009, 09:48:27 pm »

Heh, forgot about the modulus thing

This question is from the Heinemann Book and they seem to just take the positive

The inital reason I posted this though was because the book's answer was

$y = \frac{15e^{6x}-3}{1+5e^{6x}}$

With other similar questions, my answers were different to the book's but I can rearrange it fit the book's answer. I can't rearrange it in this case

TrueTears · « **Reply #6 on:** July 17, 2009, 10:27:05 pm »

Quote from: monokekie on July 17, 2009, 10:24:33 pm

in that case, you should try
Quote from: dummy on July 17, 2009, 09:27:16 pm
$9-x^2=(3-x)(3+x)$

that's where i work differently. u might be right
seems like you stumbled a little on a minor mistake, try this and rearrange it, the answer should be correct. (i got the answer stated in the book)

It is the same because you get the mod in the log, take out the - and it's same.

hyperblade01 · « **Reply #7 on:** July 22, 2009, 05:26:40 pm »

Looks alright at first, but then i screwed up..

If anyone is able to show full working that would be great

(might be better to do by hand and upload)

$\int \frac{1}{x^3+8} dx$

TrueTears · « **Reply #8 on:** July 22, 2009, 06:06:49 pm »

lol this question took me 3 pages of working I will scan up soon...

TrueTears · « **Reply #9 on:** July 22, 2009, 06:46:00 pm »

$\int \frac{1}{x^3+8} dx = \int \frac{1}{(x+2)(x^2-2x+4)} dx$

$\frac{1}{(x+2)(x^2-2x+4)} = \frac{A}{x+2} + \frac{Bx+C}{x^2-2x+4}$

$1 = A(x^2-2x+4) + (Bx+C)(x+2)$

$1 = Ax^2 - 2Ax + 4A + Bx^2 + 2Bx + Cx + 2C$

$1 = x^2(A+B) + x(-2A+2B+C) + 2C + 4A$

$A+B = 0$ ...1 $\implies B = -A$

$-2A+2B+C = 0$ ...2

$2C + 4A = 1$ ...3 $\implies C = \frac{1}{2} - 2A$

Sub all into 2.

$-2A-2A+\frac{1}{2}-2A = 0$

$A = \frac{1}{12}$

$\therefore B = \frac{-1}{12} and C = \frac{1}{3}$

$So \frac{1}{x^3+8} = \frac{1}{12(x+2)} + \frac{\frac{-1}{12}x + \frac{1}{3}}{x^2-2x+4}$

$= \frac{1}{12(x+2)} + \frac{\frac{4-x}{12}}{x^2-2x+4} = \frac{1}{12(x+2)} + \frac{-x+4}{12(x^2-2x+4)}$

$= \frac{1}{12(x+2)} - \frac{x}{12(x^2-2x+4)} + \frac{1}{3(x^2-2x+4)}$

So lets integrate step by step.

First consider:

$\int \frac{1}{12(x+2)} dx = \frac{1}{12}log_e|x+2|$ ...[1]

Second consider:

$\int \frac{1}{3(x^2-2x+4)} dx = \frac{1}{3} \int \frac{1}{x^2-2x+1-1+4} dx$

$= \frac{1}{3} \int \frac{1}{(x-1)^2+3} dx$

Let $u = x - 1 \therefore du = dx$

$\frac{1}{3} \int \frac{1}{u^2 + 3} du = \frac{1}{3} \times \frac{1}{\sqrt{3}} \int \frac{\sqrt{3}}{u^2+(\sqrt{3})^2} du$

$= \frac{1}{3\sqrt{3}} tan^{-1}(\frac{u}{\sqrt{3}})$

$= \frac{1}{3\sqrt{3}} tan^{-1}(\frac{x-1}{\sqrt{3}})$ ...[2]

Third consider:

$\int - \frac{x}{12(x^2-2x+4)} dx = \frac{-1}{12} \int \frac{x}{x^2-2x+4} dx$

From [2] we know that $x^2-2x+4 = (x-1)^2+3$

So $\frac{-1}{12} \int \frac{x}{(x-1)^2+3} dx$

Let $u = x - 1 \therefore du = dx$ and $x = u+1$

$\frac{-1}{12} \int \frac{u+1}{u^2+3} du = \frac{-1}{12}[ \int \frac{u}{u^2+3} du + \int \frac{1}{u^2+3} du]$

First let's look at $\int \frac{u}{u^2+3} du$

Let $w = u^2 + 3$

$\frac{dw}{du} = 2u \therefore dw = 2udu$

$\frac{1}{2} \int \frac{2u}{w} du = \frac{1}{2} \int \frac{1}{w} dw$

$= \frac{1}{2} log_e|w| = \frac{1}{2} log_e|u^2+3| = \frac{1}{2} log_e|(x-1)^2+3|$

Now let's look at $\int \frac{1}{u^2+3} du$

$\frac{1}{\sqrt{3}} \int \frac{\sqrt{3}}{u^2 + (\sqrt{3})^2} du = \frac{1}{\sqrt{3}}tan^{-1}(\frac{u}{\sqrt{3}}) = \frac{1}{\sqrt{3}}tan^{-1} (\frac{x-1}{\sqrt{3}})$

$\therefore \int \frac{-x}{12(x^2-2x+4)} dx = \frac{-1}{12} [ \frac{1}{2} log_e|(x-1)^2+3| + \frac{1}{\sqrt{3}}tan^{-1} (\frac{x-1}{\sqrt{3}})]$ ...[3]

Now FINALLY let's combine [1] [2] and [3]

We get : $\frac{1}{12}log_e|x+2| + \frac{1}{3\sqrt{3}} tan^{-1}(\frac{x-1}{\sqrt{3}}) + \frac{-1}{12} [ \frac{1}{2} log_e|(x-1)^2+3| + \frac{1}{\sqrt{3}}tan^{-1} (\frac{x-1}{\sqrt{3}})]$

[Excuse me for any typos or anything I haven't latexed this long before =S]

dcc · « **Reply #10 on:** July 22, 2009, 06:50:41 pm »

kamil9876 · « **Reply #11 on:** July 22, 2009, 06:50:51 pm »

dcc · « **Reply #12 on:** July 22, 2009, 06:52:04 pm »

Z is cooler then C

kamil9876 · « **Reply #13 on:** July 22, 2009, 07:07:25 pm »

i agree, i was refering to TT's post but you bet me

TrueTears · « **Reply #14 on:** July 22, 2009, 07:10:50 pm »

Quote from: kamil9876 on July 22, 2009, 07:07:25 pm

i agree, i was refering to TT's post but you bet me

bet you on what?

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