(2x-1)dx+(3y+a)dy=0

Lösung

(2 x - 1) d x + (3 y + a) d y = 0

y = \frac{- a + a ^{2} - 6 ( x ^{2} - x - c _{1} )}{3}, y = - \frac{a + a ^{2} - 6 ( x ^{2} - x - c _{1} )}{3}

Schritte zur Lösung

(2 x - 1) d x + (3 y + a) d y = 0

Solve separable ODE:

y = \frac{- a + a ^{2} - 6 ( x ^{2} - x - c _{1} )}{3}, y = - \frac{a + a ^{2} - 6 ( x ^{2} - x - c _{1} )}{3}

y = \frac{- a + a ^{2} - 6 ( x ^{2} - x - c _{1} )}{3}, y = - \frac{a + a ^{2} - 6 ( x ^{2} - x - c _{1} )}{3}

y^{^{'}} = \frac{7 x ^{2} - 1}{7 + 9 y}

\frac{d y}{d t} = e^{t + y}

x y^{^{'}} + y = 1

3 t \frac{d y}{d t} = y cos (t)

y^{^{'}} = 4 x y^{2}