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November 9th, 2016, 03:12 AM   #11
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Quote:
 Originally Posted by Mrkukas I get 1/2ln(2x+1)=ln(t/(t+1))+ln(C), but what should I do then?
$$\frac12 \ln{|2x+1|}=\ln{\left| \frac{t}{t+1} \right|}+\ln{C_1} \quad (C_1 \gt 0)$$

Note the absolute values that you should have from your integral. Combine the terms on the right and multiply by 2.

$$\ln{|2x+1|}=2\ln{\left| \frac{C_1 t}{t+1} \right|} \quad (C_1 \gt 0)$$

Use another logarithmic identity and eliminate the logarithms.

$$|2x+1|= \left| \frac{C_1 t}{t+1} \right|^2 \quad (C_1 \gt 0)$$

Bringing the exponent inside the absolute value bars, the right-hand side is now guaranteed to be positive, so they are redundant.

$$|2x+1|= \left( \frac{C_1 t}{t+1} \right)^2 = \frac{C_1^2 t^2}{(t+1)^2} \quad (C_1 \gt 0)$$

Now we remove the absolute value signs from the left-hand side.

$$2x+1 = \pm \frac{C_1^2 t^2}{(t+1)^2} \quad (C_1 \gt 0)$$

We can absorb the $\pm$ into the constant, setting $C=\pm C_1^2 \ne 0$.

$$2x+1 = \frac{C t^2}{(t+1)^2} \quad (C \ne 0)$$

Finally, since you divided by $(2x+1)$ when separating your variables, this work assumes that $2x+1\ne 0$. So now we put $x=-\frac12$, a constant function with a derivative of zero, into the original equation. Finding that the equation is satisfied, we note that if $2x+1=0$, the right-hand side of our working must also be zero, which we can achieve by allowing $C=0$. Thus

$$2x+1 = \frac{C t^2}{(t+1)^2} \quad (\text{for all C})$$

Last edited by skipjack; November 20th, 2016 at 07:18 AM.

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