2x^6y^'+9x^8y^4=0

Lösung

2 x^{6} y^{^{'}} + 9 x^{8} y^{4} = 0

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

Schritte zur Lösung

2 x^{6} y^{'} (x) + 9 x^{8} y^{4} = 0

Solve separable ODE:

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

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

(1 + x) y^{^{'}} = 4 y, y (0) = 2

x^{2} d x - (x^{3} y^{2} + 3 y^{2}) d y = 0

\frac{d v}{d t} = 3 v

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

y^{^{'}} + 3 x^{2} y = 5 x^{2}