Question

y'+xy=xy^2

Answer 1

Bernoulli equation?!

Answer 2

heh, looks like it :)

Answer 3

I thought we let \(w=y^{-2}\)

Answer 4

Let \(u = y^{-1}\)

Answer 5

Callisto yes its bernoulli equation

Answer 6

\[\frac{du}{dx} = (1-2) y^{-2} \frac{dy}{dx}=-\frac{y'}{y^2}\]
\[y'+xy=xy^2\]\[\frac{y'}{y^2} + \frac{x}{y} = x\]\[-u'+xu = x\]\[u'-xu = -x\]Integrating factor will do.

Answer 7

Can you solve the DE from here?

Answer 8

i cant because i dont understand it please help me Callisto

Answer 9

That's odd...

Answer 10

Where don't you understand?

Answer 11

i cant solve DE

Answer 12

couldn't you just separate variables
\[\frac{dy}{dx} = x(y^{2} -y)\]
\[\int\limits_{}^{}\frac{dy}{y^{2}-y} = \int\limits x dx\]
integrate LHS using partial fractions ...

Answer 13

Smart cow XD

Answer 14

@reza_anwar , can you solve it from here or should i continue?

Answer 15

plese continue it dumbcow

Answer 16

ok
\[\frac{1}{y(y-1)} = \frac{A}{y-1}+\frac{B}{y}\]
such that
Ay +B(y-1) = 1
(A+B)y -B = 1

--> A+B = 0
--> B = -1
--> A = 1
\[\rightarrow \int\limits \frac{dy}{y-1} -\frac{dy}{y} = \int\limits x dx\]
\[\ln (y-1) - \ln(y) = \frac{x^{2}}{2} +C\]
\[\ln( \frac{y-1}{y}) = \frac{x^{2}}{2}+C\]
\[\frac{y-1}{y} = k e^{x^{2}/2}\] (k = e^C)
rearrange the terms
\[y - y(ke^{x^{2}/2}) = 1\]
\[y = \frac{1}{1-ke^{x^{2}/2}}\]

Answer 17

@dumbcow thanks a lot

Answer 18

yw