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I have the following problem:

$(t+2)dx=2x^2dt$

First I divide both sides by $t+2$ to get: $dx = \frac {2x^2}{t+2}\,dt $ Then, divide by $2x^2$ to gey: $\frac{dx}{2x^2}=\frac{dt}{t+2}$ This will end up to: $\int \frac1{2x^2}dx=\int\frac{dt}{t+2}$

From now on I am not sure how to continue! I ended up having this equation: $\frac 1 5 x^3 = \ln (t+2)+c$

I need to find $x(t)$ now. Can somone help please?

update This is how I got $\frac 1{5} x^3$: I said because $\int \frac 1{2x^2}dx$ is $\frac 12 \int x^-2$

isnt it right?

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    Well yeah :) Make sure you ca$n$ duplicate Arturo's methodical treatment of the "power rule" for integrals; it's not a weakness you can afford to have!2012-05-03

2 Answers 2

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Note the first: When you "divide by $2x^2$", you have to be careful. I'm assuming $x$ is a function of $t$; you can only divide by $2x^2$ if $x^2$ is not the constant function $0$. You need to make a note of this, and/or verify whether $x=0$ is a solution to the equation. As it happens, $x=0$ is a solution, because then $dx = 0$ and $x^2=0$, so the original equation is satisfied.

(It's important not to lose sight of these "special solutions").

Note the second: $\int\frac{1}{2x^2}\,dx = \int\frac{1}{2}x^{-2}\,dx = \frac{1}{2}\int x^{-2}\,dx = \frac{1}{2}\left(\frac{1}{-1}x^{-2+1}\right)+ C = -\frac{1}{2x}+C.$

Note the third: After integrating, just solve for $x$.

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It's correct till this step:

$\int \frac1{2x^2}dx=\int\frac{dt}{t+2} $

Where

$\int \frac1{2x^2}dx=\frac{-1}{2x}$

And

$\int\frac{dt}{t+2} = \ln(t+2)$

(As you said)

Therefore,

$\frac{-1}{2x}= \ln(t+2)$

And then:

$x(t)=\frac{-1}{2 \ln(t+2)}$