again i tried
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Try the same trick as on the other post: and integrate.
we dont have a single dy /dx member here
we have many differentiated members on both side
You have .
On the RHS make the substitution so that so that your DE becomes .
Now integrating both sides gives , where , where .
You can integrate both sides with respect to now to find . You will need to use partial fractions.
From my previous post:
On both sides, you see, the derivative of the denominator appears in the numerator. Whenever that happens, you can do a u substitution, and you'll get du/u, which integrates to ln|u|. See how that works?
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