Question

find the general solution of the differential equation y'=4+2x+2y+xy

Answer 1

y'-y(2+x)=2x+4

Answer 2

check my algebra

Answer 3

from here it is just First Order Linear Differential Equation

Answer 4

i don't see how u got it to that form

Answer 5

yes i do sorry

Answer 6

find integrating factor

\[e^{\int 2+x}\]

Answer 7

dashini help

Answer 8

yea i'm there and just got that

Answer 9

\[y(x) = c_1 e ^{((x^2/2)+2 x)}-2\]

Answer 10

This is your integrating factor
\[e^{\left(2x+\frac{x^2}{2}\right)}\]

Answer 11

right

Answer 12

now put that back into the original equation and solve it for y correct?

Answer 13

yes

Answer 14

okay great...thank you

Answer 15

\[\left(y e^{\left(2x+\frac{x^2}{2}\right)}\right)'=(2x+4)\left( e^{\left(2x+\frac{x^2}{2}\right)}\right)\]

Answer 16

\[\left(y e^{\left(2x+\frac{x^2}{2}\right)}\right)=\int (2x+4)\left( e^{\left(2x+\frac{x^2}{2}\right)}\right) \text{dx}\]

Answer 17

\[\left(y e^{\left(2x+\frac{x^2}{2}\right)}\right) =2 e^{\frac{1}{2} x (4+x)}+C\]

Answer 18

Though it is not obvious you can factor
4 + 2x +2y +xy
into x(y+2) +2(y+2) = (x+2)(y+2)
and now you have a separable equation
dy/(y+2) = (x+2)dx

Answer 19

i like that way better :)

Answer 20

\[y'=4+2x+2y+xy=(x+2)(y+2)\]\[\frac{y'}{y+2}=(\ln(y+2))'=x+2\]\[\ln{(y+2)}=\frac{x^2}{2}+2x+C\]\[y=e^{x^2/2+2x+C}-2=C_1e^{x^2/2+2x}-2\]

Answer 21

The first way is more general.

Answer 22

yeah, much easier using separable equation

Answer 23

now with the condition of y(0) = 0 does this change everything?

Answer 24

nope, you just get value for C