15x^2y+3x^3y^'-24y=0

Lösung

15 x^{2} y + 3 x^{3} y^{^{'}} - 24 y = 0

y = \frac{e ^{\frac{- 4 + c _{1} x ^{2}}{x ^{2}}}}{x ^{5}}

Schritte zur Lösung

15 x^{2} y + 3 x^{3} y^{'} (x) - 24 y = 0

Solve separable ODE:

y = \frac{e ^{\frac{- 4 + c _{1} x ^{2}}{x ^{2}}}}{x ^{5}}

y = \frac{e ^{\frac{- 4 + c _{1} x ^{2}}{x ^{2}}}}{x ^{5}}

\frac{d y}{d x} = x e^{y}, y (0) = 1

1 - x^{2} \frac{d y}{d x} = x + x y^{2}

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

\frac{d y}{d x} = - 5 y, y = 7, x = 0

\frac{d y}{d x} + 2 y^{2} = 2 y