Question

differential equation & linear algebra equation.
I need to know how to use Bernoulli Equation to solve for xy' + y + x^2y^2e^x

Answer 1

i want to know the step by step.

Answer 2

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

Answer 3

uh... step by steps?

Answer 4

http://en.wikipedia.org/wiki/Bernoulli_differential_equation

\[z=y^{1-2}=y^{-1}\]

Answer 5

.... okay thanks

Answer 6

lol..wikipedia

Answer 7

oh! can you help?!

Answer 8

i'll let @mukushla do his thing

Answer 9

....

Answer 10

his using wikipedia...

Answer 11

\[z'=-\frac{y'}{y^2}\]\[xy'+y=x^2e^xy^2\]divide by y^2\[x\frac{y'}{y^2}+\frac{1}{y}=x^2e^x\]now u have\[-xz'+z=x^2e^x\]

Answer 12

this is a first order linear diff equation solve it for z

Answer 13

okay thank you for the help ~.~

Answer 14

welcome

Answer 15

do you know the form of Bernoulli's equation @adunb8 ?

Answer 16

well know i looked at it it looks like its \[y' + p(x)y = q(x)y^n\]

Answer 17

right

Answer 18

so your first step should be to transform your equation into that form

Answer 19

yes i made it into \[y' + x^{-1} y = xy^2e^x\]

Answer 20

wait....what's the original equation?

Answer 21

\[xy' + y + x ^{2}y ^{2}e ^{x}\]

Answer 22

= 0