Question

solve the given differential equation by using appropriate solutions

x*dx+(y-2x)dy=0

Answer 1

are bernoulli's homogenious

Answer 2

wasn't Bernoulli's exactly ... it was first order linear differential equation.

Answer 3

I cant view the whole link

Answer 4

http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&ved=0CDoQFjAC&url=http%3A%2F%2Fcollege.cengage.com%2Fmathematics%2Flarson%2Fcalculus_early%2F3e%2Fshared%2Fchapter15%2Fclc7eap1502.pdf&ei=WwyCT8H8Iou3rAf6tI3iBQ&usg=AFQjCNHxp2mjiIqPx7M-Fg4bsz2C9Zib2Q&sig2=HEZTU6WvveVgecGv0gntDg

Answer 5

I know I have to use substitution, but I don't know what to substitute, the answer is attatched below

Answer 6

my previous answer was also incorrect, should have seen the equation more clearly,

Answer 7

change y=vx, dy/dx = xdv/dx + v

Answer 8

also cancel out, x's on the ratio, ... then use variable separation method.

Answer 9

i dont get it could you walk me through it

Answer 10

dy/dx+x/(y-2x)=0

put y = vx
then you will have
dy/dx = xdv/dx + v

or
xdv/dx + v +x/(vx-2x) = 0

now solve using variable separation method.

Answer 11

wouldnt it be (y-2x)/x ?

Answer 12

?? y is changed

Answer 13

in to vx => cancel out that x's on second term

Answer 14

I assume I can integrate now?

Answer 15

after separating variables, v and x

Answer 16

this seems good?

Answer 17

before the integration

Answer 18

yeah .. must be something like that.

Answer 19

im not getting teh answer

Answer 20

that they have in book

Answer 21

change v back to y/x

Answer 22

this is a homogeneous differential equation, you just need to substitute\[y = zx\]and then the it will become a separable equation

Answer 23

can some one solve this for me because I have abosolutely no clue

Answer 24

is it x dx or x*dx???

Answer 25

x dx

Answer 26

ok let me if i can help u

Answer 27

\[xdx+\left( y-2x \right)dy=0\]\[\frac{dy}{dx}=\frac{-x}{y-2x}\]let\[y=zx\]\[\frac{dy}{dx}=\frac{dz}{dx}x + z\]and then substitute to the equation\[\frac{dz}{dx}x+z=\frac{-x}{zx - 2x}\]\[\frac{dz}{dx}x=\frac{-z^{2}+2z-1}{z+2}\]\[\frac{z+2}{-z^{2} + 2z - 1}dz=\frac{dx}{x}\]now you just need to integrate both sides