y^'=y^{1/5}

Lösung

y^{^{'}} = y^{\frac{1}{5}}

y = \frac{2 \cdot 4 4 t ^{5} + c _{1}^{5} + 5 c _{1} t ^{4} + 10 c _{1}^{2} t ^{3} + 10 c _{1}^{3} t ^{2} + 5 c _{1}^{4} t}{5 4 5}, y = - \frac{2 \cdot 4 4 t ^{5} + c _{1}^{5} + 5 c _{1} t ^{4} + 10 c _{1}^{2} t ^{3} + 10 c _{1}^{3} t ^{2} + 5 c _{1}^{4} t}{5 4 5}

Schritte zur Lösung

y^{'} (t) = y^{\frac{1}{5}}

Solve separable ODE:

y = \frac{2 \cdot 4 4 t ^{5} + c _{1}^{5} + 5 c _{1} t ^{4} + 10 c _{1}^{2} t ^{3} + 10 c _{1}^{3} t ^{2} + 5 c _{1}^{4} t}{5 4 5}, y = - \frac{2 \cdot 4 4 t ^{5} + c _{1}^{5} + 5 c _{1} t ^{4} + 10 c _{1}^{2} t ^{3} + 10 c _{1}^{3} t ^{2} + 5 c _{1}^{4} t}{5 4 5}

y = \frac{2 \cdot 4 4 t ^{5} + c _{1}^{5} + 5 c _{1} t ^{4} + 10 c _{1}^{2} t ^{3} + 10 c _{1}^{3} t ^{2} + 5 c _{1}^{4} t}{5 4 5}, y = - \frac{2 \cdot 4 4 t ^{5} + c _{1}^{5} + 5 c _{1} t ^{4} + 10 c _{1}^{2} t ^{3} + 10 c _{1}^{3} t ^{2} + 5 c _{1}^{4} t}{5 4 5}

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

\frac{d y}{d x} y = arccos (\frac{1}{x})

y^{^{'}} - \frac{7}{5} y = 0

y^{^{'}} + y = y (t e^{t^{2}} + 1) t, y (0) = 1

\frac{d y}{d x} y = (x^{x - 3})