
Differential Equations Ordinary and Partial Differential Equations Math Forum 
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November 9th, 2016, 03:12 AM  #11 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,973 Thanks: 2296 Math Focus: Mainly analysis and algebra  $$\frac12 \ln{2x+1}=\ln{\left \frac{t}{t+1} \right}+\ln{C_1} \quad (C_1 \gt 0)$$ 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 righthand 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 lefthand 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 righthand 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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