\[x^3y'+3y^2=xy^2\]

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\[\frac{dy}{dx} \cdot x^3=y^2(x-3)\]
\[\frac{1}{y^2} dy=\frac{x-3}{x^3} dx\]
integrate both sides

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Other answers:

\[\int\limits_{}^{}y^{-2} dy =\int\limits_{}^{}(x^{-2}-3x^{-3}) dx\]
\[\frac{y^{-2+1}}{-2+1}=\frac{x^{-2+1}}{-2+1}-3 \cdot \frac{x^{-3+1}}{-3+1}+C\]
\[-\frac{1}{y}=\frac{-1}{x}-3\frac{1}{-2x^{2}}+C\]
\[\frac{-1}{y}=\frac{-1}{x}+\frac{3}{2x^2}+C\]
it was just separation by variables
thankyou myininaya, i could not see that is was just separation of variable. but the answer in my text has arctans and lns, i guess i need to rearrange a negative before integration
that is the right problem right?
thats weird to me that it would have arctans and lns
im sorry apparently i dont know how to read, your answer is perfect
lol
:)
goodnight unkle great job on explaining fourier to matt
it is 3 in the afternoon to me, but good night anyway, thanks

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