y^'=(x+xy^2)/(4y),y(1)=0

Lösung

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

y = e^{\frac{x ^{2} - 1}{4}} - 1, y = - e^{\frac{x ^{2} - 1}{4}} - 1

Schritte zur Lösung

y^{'} (x) = \frac{x + x y ^{2}}{4 y}

Solve separable ODE:

y = e^{\frac{x ^{2} - 1}{4}} - 1, y = - e^{\frac{x ^{2} - 1}{4}} - 1

y = e^{\frac{x ^{2} - 1}{4}} - 1, y = - e^{\frac{x ^{2} - 1}{4}} - 1

y^{^{'}} - 9 x^{2} y + y = 0

y^{^{'}} = x (y - 1) - 2 y + 2

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

y^{^{'}} - 6 x y = 0

y^{^{'}} + y \cdot cos (t) = - cos (t)