/sci/ - Science & Math

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no mistake in your steps above, try calculating the integrals again.

>>4801232
Your steps are wrong. But your end result
<div class="math">y=\frac {2x}{1 -2xC}</div>
is the solution.

>>4801241
Alright, will do.

>>4801246

Ah? Care to explain?

<span class="math">\displaystyle{\frac{dy}{dx} = \frac{y^2}{2x^2} }[/spoiler]

<span class="math">\displaystyle{\frac{dy}{y^2} = \frac{dx}{2x^2} }[/spoiler]

<span class="math">\displaystyle{\int \frac{dy}{y^2} = \int \frac{dx}{2x^2} }[/spoiler]

<span class="math">\displaystyle{ -\frac{1}{y} = -\frac{2}{x} + C}[/spoiler]

<span class="math">\displaystyle{ y = - \frac{1}{\frac{2}{x} + C}}[/spoiler]

>>4801253
Your are already dividing by (2x^2) which can be 0 in your first step and I will not comment on the 'multiplication with dx'.

i got the same result OP. Must be a mistype on the textbooks publishers side

>>4801259
no

i got y=2x/(1-cx)

>>4801259
line 4 and 5 the 2 is in the denominator

<span class="math">\displaystyle{ \frac{-1}{y} = \frac{-1}{2x} + C}[/spoiler]


<span class="math">\displaystyle{ y = \frac{-1}{\frac{1}{2x} + C}}[/spoiler]

>>4801260
shut the fuck up

>>4801258
Even if that were the case, Wolfram says otherwise.

>>4801270
what does the book say

>>4801274

Interestingly,

<span class="math"> \frac{2x}{Cx +1}[/spoiler]

>>4801267
another mistake on line 2, the minus sign

<span class="math">\displaystyle{ y = \frac{-1}{\frac{-1}{2x} + C}} [/spoiler]

sorry for all the mistakes, i'm sleepy

>>4801281
You are fucking retarded. A constant is a constant no matter if you write it as C or -2C.

>>4801281
there's a typo on the 'Cx' it should be '2Cx',

this is the answer
>>4801283

follow the previous steps, there is no big deal here, it's one of the simplest first order linear differential equations there are

>>4801289
Thanks. Oh textbook, how could you lie to me.

>>4801286
or it could also be this. since it is not stated what C is, you could call it <span class="math">C = 2C_1[/spoiler], where C_1 is the original constant from the antiderivaties

>>4801294
Pardon. I shoud've written <span class="math">C_1[/spoiler].

>>4801286

<span class="math">\int\frac{2\frac{dy(x)}{dx}}{y(x)^2}dx = \int\frac{1}{x^2}dx[/spoiler]

<span class="math">-\frac{2}{y(x)} = -\frac{1}{x} + C[/spoiler]

<span class="math">y(x) = -\frac{2x}{Cx-1}[/spoiler]

>>4801298

Ah yeah, that. When I checked the way Wolfram calculated it, I saw the

<span class="math"> \int\frac{2\frac{dy(x)}{dx}}{y(x)^2}dx [/spoiler].

What exactly happened there?

>>4801306
<span class="math">y' = \frac{dy(x)}{dx}[/spoiler]

>>4801306
http://en.wikipedia.org/wiki/Integration_by_substitution

>>4801314
Duh. Silly question, I admit.

>>4801259

How do you get TeX equations to display in a larger size like this?

>>4801328
\displaystyle{ YOUR SHIT HERE }

>>4801332

<span class="math">\displaystyle{\frac{Thanks}{Anon}}[/spoiler]