Question

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

Answer 1

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

Answer 2

\[\frac{1}{y^2} dy=\frac{x-3}{x^3} dx\]

Answer 3

integrate both sides

Answer 4

\[\int\limits_{}^{}y^{-2} dy =\int\limits_{}^{}(x^{-2}-3x^{-3}) dx\]

Answer 5

\[\frac{y^{-2+1}}{-2+1}=\frac{x^{-2+1}}{-2+1}-3 \cdot \frac{x^{-3+1}}{-3+1}+C\]

Answer 6

\[-\frac{1}{y}=\frac{-1}{x}-3\frac{1}{-2x^{2}}+C\]

Answer 7

\[\frac{-1}{y}=\frac{-1}{x}+\frac{3}{2x^2}+C\]

Answer 8

it was just separation by variables

Answer 9

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

Answer 10

that is the right problem right?

Answer 11

thats weird to me that it would have arctans and lns

Answer 12

im sorry apparently i dont know how to read,
your answer is perfect

Answer 13

lol

Answer 14

:)

Answer 15

goodnight unkle
great job on explaining fourier to matt

Answer 16

it is 3 in the afternoon to me,
but good night anyway,
thanks