y' + y = y^-2 using bernoulli's equation
bernoulli's equation says that \[u=y^{1-n}\]
go on
\[u=y^{1-2}\] \[u=y^{-1}\]
solve for y and take the derivative in respects to x (using the chain rule)
\[\frac{dy}{dx}=\frac{du}{dx}*\frac{dy}{du}\]
can you do this and tell me what you et
is the answer must be y^3 = 1+Ce^-3x??
in really not good at ODE..i just need help..
so did you use the above and substitute in? what did you get for your linear equation in u
i did not sole that i just guess that from the choices
what do you think is the right solution for that?
i can't just give you the answer. You need to do some of the work also. I'm sorry
if you give me the right answer i can review for that and know what is wrong from my answer.
sorry but i think \[u=y^{1-(-2)}=y^3\]
so your saying that it is y^3 = 1+Ce^-2x??
ah yes that is correct @mukushla
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