Find the general solution of:y^'=(t-2y)/(2t+y)

Answer 1

y' = (t - 2y)/(2t + y). This is homoginius equation: y = vt; y' = v't + v;

v't + v = (t - 2tv)/(2t + tv); v't = (1 - 2v)/(2 + v) - v; v't = (1 - 2v - 2v - v²)/(2 + v);

(2 + v)/(1 - 4v - v²) = dt/t; ∫(2 + v)/(1 - 4v - v²)dv = ∫dt/t; - 1/2lnI1 - 4v - v²I = lnItI - 1/2lnC

Iv² + 4v - 1I = CItI^-2. v = y/t; Iy²/t² + 4y/t - 1I = CItI^-2 or Iy² + 4yt - t²I = C