Question

Anyone can help me in this bernoulli's equation

Answer 1

(3siny -5x)dx +2x^2 cotydy=0

Answer 2

@jiteshmeghwal9

Answer 3

i found in some source now ^.^

Answer 4

file:///C:/Users/User/Downloads/ENGIANA+QUIZ+1.pdf

Answer 5

Might need a substitution first. Maybe something like \(\sin y=u\)? This makes \(\dfrac{\mathrm du}{\mathrm dx}=\cos y\,\dfrac{\mathrm dy}{\mathrm dx}\), or \(\dfrac{1}{\sqrt{1-u^2}}\,\dfrac{\mathrm du}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dx}\). So the ODE becomes
\[3\sin y-5x+2x^2\cot y\frac{\mathrm dy}{\mathrm dx}=0\\[3ex]
\implies3u-5x+2x^2\frac{\sqrt{1-u^2}}{u}\left(\frac{1}{\sqrt{1-u^2}}\frac{\mathrm du}{\mathrm dx}\right)=0\]or
\[\frac{2x^2}{u}\frac{\mathrm du}{\mathrm dx}=5x-3u\]But that doesn't seem right to me, either because I made a mistake somewhere or this equation is indeed not a Bernoullie equation.

Answer 6

go to this link: they got the answer already file:///C:/Users/User/Downloads/ENGIANA+QUIZ+1.pdf

Answer 7

@HolsterEmission can you help me on next problem ^.^

Answer 8

I can't access files that are on your computer :)

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