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I have the following problem:


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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How did you get $\dfrac{1}{5} x^3$ ?? –  The Chaz 2.0 May 3 '12 at 19:37
It looks like you should have $$\dfrac{-1}{2x} = \dfrac{\ln (t + 2) + c}{1}$$ (I added the "1" for effect) ... can't you cross multiply from here? –  The Chaz 2.0 May 3 '12 at 19:39
It is indeed $\frac {1}{2} \int \frac{1}{x^2}$ which is $\frac {1}{2} (\frac{-1}{x}+C)$. –  Gigili May 3 '12 at 19:53
Sean, the 1/5 is still puzzling to me... you should divide by (exponent + 1), but it looks like you're subtracting (exponent - 1) from the denominator...??? –  The Chaz 2.0 May 3 '12 at 19:53
@The Chaz Yeah I was doing it wrong :P –  Sean87 May 3 '12 at 19:54

2 Answers 2

up vote 5 down vote accepted

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:

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


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


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

(As you said)


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

And then:

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

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