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Originally Posted by

**jescavez** What is that technique called?

I don't think it has a name. :D

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why in:

$\displaystyle

zy'+y=-\frac{1}{2} \implies y'=-\frac{1}{z}\left(y+\frac{1}{2}\right) \Longleftrightarrow \left( y+\frac{1}{2}\right)'=-\frac{1}{z}\left( y+\frac{1}{2}\right)

$

does the implied bit work?

I don't understand what you mean :(

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and I'm not quite sure how you get to:

$\displaystyle

y+\frac{1}{2}=\exp\left(\int -\frac{\mathrm{d}u}{u}\right)

$

You should take a look at this page : first-order ODE

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(And last question, hopefully, sorry!)

There is no need to apologize each time you ask a question. :)

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For this bit:

$\displaystyle

\begin{cases}

f(1)=0\\

f(e^2)=0

\end{cases}

\implies\begin{cases}

1\ln 1-\frac{1}{2}\cdot 1+B=0\\

A\ln e^2-\frac{1}{2}e^2+B=0

\end{cases}

\implies

\begin{cases}

B=\displaystyle\frac{1}{2}\\

A=\displaystyle\frac{e^2-1}{4{\color{red}\ln e}}

\end{cases}

$

we are adding the solutions right to get to final equation? then what happens to the $\displaystyle -\frac{1}{2}e^2+B$ part of $\displaystyle f(e^2)$? we do not seem to use it?

We do use this term :

$\displaystyle

\begin{aligned}

A\ln e^2-\frac{1}{2}e^2+B=0

&\implies A\ln e^2-\frac{1}{2}e^2+\frac{1}{2}=0\\

&\implies 2A\ln e = \frac{e^2-1}{2}\\

&\implies A=\frac{e^2-1}{4\ln e}\\

\end{aligned}$

If you want to do this by subtracting one equation to the other :

$\displaystyle

\begin{aligned}

\begin{cases}

1\ln 1-\frac{1}{2}\cdot 1+B=0 & (1)\\

A\ln e^2-\frac{1}{2}e^2+B=0 &(2)

\end{cases} &\implies (2)-(1) : A\ln e^2-\frac{1}{2}e^2+\frac{1}{2}=0\\

&\implies A=\frac{\frac{1}{2}e^2-\frac{1}{2}}{2\ln e}\\

&\implies A=\frac{e^2-1}{4\ln e}

\end{aligned}$