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Mathematics 23 Online
OpenStudy (anonymous):

y' + y = y^-2 using bernoulli's equation

OpenStudy (anonymous):

bernoulli's equation says that \[u=y^{1-n}\]

OpenStudy (anonymous):

go on

OpenStudy (anonymous):

\[u=y^{1-2}\] \[u=y^{-1}\]

OpenStudy (anonymous):

solve for y and take the derivative in respects to x (using the chain rule)

OpenStudy (anonymous):

\[\frac{dy}{dx}=\frac{du}{dx}*\frac{dy}{du}\]

OpenStudy (anonymous):

can you do this and tell me what you et

OpenStudy (anonymous):

is the answer must be y^3 = 1+Ce^-3x??

OpenStudy (anonymous):

in really not good at ODE..i just need help..

OpenStudy (anonymous):

so did you use the above and substitute in? what did you get for your linear equation in u

OpenStudy (anonymous):

i did not sole that i just guess that from the choices

OpenStudy (anonymous):

what do you think is the right solution for that?

OpenStudy (anonymous):

i can't just give you the answer. You need to do some of the work also. I'm sorry

OpenStudy (anonymous):

if you give me the right answer i can review for that and know what is wrong from my answer.

OpenStudy (anonymous):

sorry but i think \[u=y^{1-(-2)}=y^3\]

OpenStudy (anonymous):

so your saying that it is y^3 = 1+Ce^-2x??

OpenStudy (anonymous):

ah yes that is correct @mukushla

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