EMRE C.
asked • 11/14/21MATH CALCULUS QUESTİONS
find the ordinary differential equation solution
(dy)/(dx) = (x - 2y + 6)/(2x + y + 2)
y'=-8xy² + 4x (4x+1)y-(8x3+4x²-1)1, y=x-custom solution
(3x + 2y2)dx + 2xydy = 0
(sec² y) y'-3 tany = - 1
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1 Expert Answer
MUHAMMAD I. answered • 02/10/22
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For the equation (x - 2y + 2)dx + (2x + y - 6)dy =0 take x = U+2 , y = V+2.
The equation becomes homogeneous as ( U - 2V )dU + ( 2U + V )dV =0 . Then
let W = V/U and obtain ( 1–2W)dU + (2+ W)(WdU + UdW) =0.This can be separated
- dU/U = ( W + 2 )dW/(W^2 + 1) = 2WdW/2(W^2 + 1) + 2dW/(W^2+1) . Integrating
(1/2)ln(W^2+1) + 2arctanW = - lnU + lnC . Obtain
ln[U^2(W^2+1)/K] = -4arctanW , K = C^2. Then
( V^2 + U^2) = Ke^(-4arctanV/U). Recall U = x-2 , V = y-2 and obtain
the solution in term of old variables.
Luke J.
MUHAMMAD I., this approach looks very interesting and will try to work thru it on my own to verify your method...but could you go about it of the form M(x,y) dx + N(x,y) dy = 0 and check that it's exact (dM/dy = dN/dx except the 'd' s are partial derivatives, didn't have enough time to search for that) and go about it that way?
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02/10/22
Luke J.
Also, you have a typo mistake on line 1, (x - 2y + 2)dx + (2x + y - 6)dy = 0 should actually be ( 2y - x - 6 ) dx + ( 2x + y + 2 )dy = 0; i believe there may have been an error in your cross multiplication of some sort
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02/10/22
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EMRE C.
(3x + 2y2)dx + 2xydy = 0 ==f(x,y)=x^2y^2+x^3+c11/14/21