10x^2y+2x^3y^'-16y=0

Lösung

10 x^{2} y + 2 x^{3} y^{^{'}} - 16 y = 0

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

Schritte zur Lösung

10 x^{2} y + 2 x^{3} y^{'} (x) - 16 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}}

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

\frac{d y}{d x} = \frac{y x + 3 x - y - 3}{x y - 2 x + 4 y - 8}

\frac{d y}{d x} = \frac{9 ( 1 + x )}{x y}

x^{^{'}} = \frac{2 x}{t + 1}

y^{^{'}} + (x + 9) y = 0