(t+2y+3)+(2t+4y-1)(dy)/(dt)=0

解

(t + 2 y + 3) + (2 t + 4 y - 1) \frac{d y}{d t} = 0

y = \frac{1 - 2 t + - 28 t + c _{1} + 1}{4}, y = \frac{1 - 2 t - - 28 t + c _{1} + 1}{4}

解答ステップ

(t + 2 y + 3) + (2 t + 4 y - 1) \frac{d y}{d t} = 0

完全 ODE を解く:

y = \frac{1 - 2 t + - 28 t + c _{1} + 1}{4}, y = \frac{1 - 2 t - - 28 t + c _{1} + 1}{4}

y = \frac{1 - 2 t + - 28 t + c _{1} + 1}{4}, y = \frac{1 - 2 t - - 28 t + c _{1} + 1}{4}