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(2x^2+xy+y^2)dx + 2x^2 dy=0 How Do I Solve This 1st Order DifEQ? Help Please!!
yup this 1st Order DifEQ
you can use y=vx (for homogeneous equations) to solve it
I don't understand how you determine if its homogeneous, separable...but I understand if its an exact equation...
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..
see the sum of the powers of x and y is uniform (here its 2 ) throughout hence homogeneous
Great! I'm Learning Something Finally!
after using y=vx and dy =xdv +vdx the resultant will become a seperable one
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
You cannot solve this using separation of variables.
Yes your right, I understand that its not an exact equation..
I have 6 problems that are due in 45 mins... I need help!!!
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
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)
***I dropped a half in front of the (y/x)^2 term, so v + xv' = -(1/2) ( 2 + v + v^2 )
I keep getting different answers though when i do that... i have the solutions just can get there while showing my work
Anyone who can help solve these problems will be a life saver!
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.