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I'm trying to find the value of a constant for $\ y(0) = 0$ in the following differential equation.

$$\ 2\ln(2x+3y-1) - {2x+3y \over 2} = 2x+3y + k$$

Of course when plugging in the values, I get $\ 2\ln(-1) = k$ which errors. When entering this into Wolfram Alpha, they suggest rearranging the equation from this format,

$$\ 2\ln(x-1) - {x \over 2} + constant$$

to

$$\ - {x \over 2} + 2\ln(1-x) + {1 \over 2} + constant$$

"Which is the equivalent for restricted x values", which indeed I have. This would leave me with a positive $\ln(1)$, which would solve my problem... but...

My question, how does this manipulation work? I've never seen this before. I don't understand how they've made that leap (or if it's even accurate).

Can anyone educate me?

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Note that for any constants $a,b,c\in\mathbb{R}$, $a\ln(b(x+c))=a\ln(b)+a\ln(x+c)=\ln(b^{a})+\ln((x+c)^{a})$

Therefore, in this case we have:

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

And using the fact that $(-1)^{2}=1$, and $\ln(1)=0$ we have:

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

And then as amWhy points out, you can have $c=\frac{1}{2}+k$ for restricted values of $x$.

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  • $\begingroup$ Brilliant! That's what I was looking for. Thank you. $\endgroup$
    – AdrianR
    Feb 28 '13 at 20:05

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