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Find all solutions of y'-xy=(x^3)(y^3).
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Its Bernoulli. Get it into the form of this. \[\LARGE \frac{dy}{dx} + P(y) = Q\]
Make sure that Q should be solely a function of x.
So do I divided the whole equation by y^3?
yep!
Then what do I do?
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sibstitute v = 1/y^2
\[\Large \frac{y'}{y^3}-\frac{x}{y^2}=(x^3)\] Let v=1/y^2
rationalllllllllllllllll
Let me work it out and I'll tell you what I got.
sure!
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Since v=1/y^2, so y=1/sqrt(v), right?
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