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 one year ago
(2x^2+xy+y^2)dx + 2x^2 dy=0
How Do I Solve This 1st Order DifEQ? Help Please!!
 one year ago
(2x^2+xy+y^2)dx + 2x^2 dy=0 How Do I Solve This 1st Order DifEQ? Help Please!!

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matricked
 one year ago
Best ResponseYou've already chosen the best response.0yup this 1st Order DifEQ

matricked
 one year ago
Best ResponseYou've already chosen the best response.0you can use y=vx (for homogeneous equations) to solve it

c1c9m9h1
 one year ago
Best ResponseYou've already chosen the best response.0I don't understand how you determine if its homogeneous, separable...but I understand if its an exact equation...

c1c9m9h1
 one year ago
Best ResponseYou've already chosen the best response.0I have 6 problems that i have to do and i have the answers but i just dont understand how to do the work to get the answer unless its an exact equation..

matricked
 one year ago
Best ResponseYou've already chosen the best response.0see the sum of the powers of x and y is uniform (here its 2 ) throughout hence homogeneous

c1c9m9h1
 one year ago
Best ResponseYou've already chosen the best response.0Great! I'm Learning Something Finally!

matricked
 one year ago
Best ResponseYou've already chosen the best response.0after using y=vx and dy =xdv +vdx the resultant will become a seperable one

c1c9m9h1
 one year ago
Best ResponseYou've already chosen the best response.0Is there a certain formula that is out there where i can solve these problems if they are separable or homogenous or any other thing out there?? I really want to learn this stuff

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0You cannot solve this using separation of variables.

c1c9m9h1
 one year ago
Best ResponseYou've already chosen the best response.0Yes your right, I understand that its not an exact equation..

c1c9m9h1
 one year ago
Best ResponseYou've already chosen the best response.0I have 6 problems that are due in 45 mins... I need help!!!

JamesJ
 one year ago
Best ResponseYou've already chosen the best response.2Let's start from the beginning (2x^2+xy+y^2)dx + 2x^2 dy=0 implies that dy/dx = (2x^2+xy+y^2)/(2x^2) and hence dy/dx = 1  (1/2)(y/x)  (y/x)^2

JamesJ
 one year ago
Best ResponseYou've already chosen the best response.2This is what is called a homogeneous equation. It is one of the meanings of the word homogeneous for differential equations. The substitution 'trick' for such equations is to write v = y/x and hence y = vx which implies y' = v + xv' Therefore we can rewrite the equation above as v + xv' =  (1 + (1/2)v + v^2) This now IS a separable equation which you can solve using standard techniques to find the function v(x). Once you have done that, substitute back v(x) = y(x)/x to solve for y(x)

JamesJ
 one year ago
Best ResponseYou've already chosen the best response.2***I dropped a half in front of the (y/x)^2 term, so v + xv' = (1/2) ( 2 + v + v^2 )

c1c9m9h1
 one year ago
Best ResponseYou've already chosen the best response.0I keep getting different answers though when i do that... i have the solutions just can get there while showing my work

c1c9m9h1
 one year ago
Best ResponseYou've already chosen the best response.0Anyone who can help solve these problems will be a life saver!

JamesJ
 one year ago
Best ResponseYou've already chosen the best response.2Well, v + xv' = (1/2) ( 2 + v + v^2 ) implies 2x v' = v^2 + 3v + 2 = (v+2)(v+1) hence separating \[ 2\int \left( \frac{A}{v+2} + \frac{B}{v+1} \right) dv = \int \frac{dx}{x} \] You take it from there.
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