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c1c9m9h1
Group Title
(2x^2+xy+y^2)dx + 2x^2 dy=0
How Do I Solve This 1st Order DifEQ? Help Please!!
 one year ago
 one year ago
c1c9m9h1 Group Title
(2x^2+xy+y^2)dx + 2x^2 dy=0 How Do I Solve This 1st Order DifEQ? Help Please!!
 one year ago
 one year ago

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

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

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

matricked Group TitleBest ResponseYou've already chosen the best response.0
then dy =xdv +vdx
 one year ago

c1c9m9h1 Group TitleBest ResponseYou've already chosen the best response.0
I 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..
 one year ago

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

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

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

c1c9m9h1 Group TitleBest ResponseYou've already chosen the best response.0
Is 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
 one year ago

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

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

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

JamesJ Group TitleBest ResponseYou've already chosen the best response.2
Let'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
 one year ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.2
This 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)
 one year ago

JamesJ Group TitleBest 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 )
 one year ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.2
Make sense?
 one year ago

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

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

JamesJ Group TitleBest ResponseYou've already chosen the best response.2
Well, 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.
 one year ago
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