MATH CALCULUS QUESTİONS

Question

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

Answer 1

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.