y^'=(1+2y^2)/(ysin(x))

Lösung

y^{^{'}} = \frac{1 + 2 y ^{2}}{y sin ( x )}

y = \frac{c _{1} tan ^{4} ( \frac{x}{2} ) - 1}{2}, y = - \frac{c _{1} tan ^{4} ( \frac{x}{2} ) - 1}{2}

Schritte zur Lösung

y^{'} (x) = \frac{1 + 2 y ^{2}}{y sin ( x )}

Solve separable ODE:

y = \frac{c _{1} tan ^{4} ( \frac{x}{2} ) - 1}{2}, y = - \frac{c _{1} tan ^{4} ( \frac{x}{2} ) - 1}{2}

y = \frac{c _{1} tan ^{4} ( \frac{x}{2} ) - 1}{2}, y = - \frac{c _{1} tan ^{4} ( \frac{x}{2} ) - 1}{2}

y^{^{'}} + 7 y + 5 x = 5 x

y d x + 2 x d y = 0

3 \frac{d y}{d x} - y = 12 e^{- 5}

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

y^{^{'}} = \frac{y ^{2}}{3 x}