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\[\int \frac{dy}{2y+1}=\int dx \\
\ 2 \ln y = x
\]

Hmm, Im having trouble remembering this.

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I need to visualize integration by substitution better.

I forgot the -1 on the previous one.

y' + xy = x
dy/dx = x(1-y)
dy/(1-y) = x dx
that can be seperated as well

The first one checks with my answer, thank you! And the second one looks great too.

Nevermind, answered my own question, I believe.

:) `1/2C` is as arbitrary as `C`

`e^C` is as arbitrary as `C` is too

so yeah we can lump `1/2e^C` by a new name `C`

im not sure if thats the question though, as you have deleted it before i completely reading...

\[y_h = c_1y_1+ c_2y_2\]That's what I was thinking of.

I see..

so we get :
\[\large y_h = \sum c_iy_i\]

\(\large y_i\)'s are the independent solutions to the homogeneous part

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