y^'=(8x-5)(y+2),y(0)=7

解答

y^{^{'}} = (8 x - 5) (y + 2), y (0) = 7

y = 9 e^{4 x^{2} - 5 x} - 2

求解步骤

y^{'} (x) = (8 x - 5) (y + 2)

求解可分离方程:

y = 9 e^{4 x^{2} - 5 x} - 2

y = 9 e^{4 x^{2} - 5 x} - 2

cos (y) y^{^{'}} = sin^{2} (t) cos (t), y (0) = \frac{π}{6}

y^{^{'}} x = cos^{2} (y)

5 x \frac{d y}{d x} + 8 y = 0

y^{^{'}} + y (2 x sin (2 x) - sin^{2} (x)) = 0

(x + 1)^{2} y^{^{'}} + 8 (x + 1) y^{^{'}} + 12 y = 0