differential equation problem

[kcoe05]

here is the problem:

(x^2 - 1) dy + x(y^2 + 5y + 4) dx = 0

Do i need to get in the form dx / dy + p(y)x = g(y) ?

Thank you so much for your time and expertise?

 

[arthur ohlsten]

[x^2-1] dy + x [ y^2+5y+4] dx=0
[x^2-1] dy = - x[y+4][y+1] dx
dy / {[y+1][y+4]} = -xdx /[x^2-1]

right side
let u=x^2-1
then du= 2x dx
or right side is
-[1/2] du/u

left side
1/{[y+1][y+4]} = A/[y+4] + B /[y+1]
multiply by y+1
1/[y+4] = B + A[y+1] /[y+4]
let y=-1
1/3= B

multiply by y+4
1/[y+1] = A + B[y+4]/[y+1]
let y=-4
-1/3 =A

right side is
[-1/3]dy/[y+4] +[1/3] dy/[y+1]

or we have du/u form

you can now integrate

Arthur

 

[kcoe05]

Thank you so much. That makes a lot more sense. I greatly appreciate your help.