(2+x)y^'=5y

Lösung

(2 + x) y^{^{'}} = 5 y

y = 32 e^{c_{1}} + 80 e^{c_{1}} x + 80 e^{c_{1}} x^{2} + 40 e^{c_{1}} x^{3} + 10 e^{c_{1}} x^{4} + e^{c_{1}} x^{5}

Schritte zur Lösung

(2 + x) y^{'} (x) = 5 y

Solve separable ODE:

y = 32 e^{c_{1}} + 80 e^{c_{1}} x + 80 e^{c_{1}} x^{2} + 40 e^{c_{1}} x^{3} + 10 e^{c_{1}} x^{4} + e^{c_{1}} x^{5}

y = 32 e^{c_{1}} + 80 e^{c_{1}} x + 80 e^{c_{1}} x^{2} + 40 e^{c_{1}} x^{3} + 10 e^{c_{1}} x^{4} + e^{c_{1}} x^{5}

\frac{\partial}{\partial x} (x + \frac{2}{y})

t a y l or \frac{1}{1 - x + \frac{x ^{2}}{2}}, 0

\frac{d}{d x} (x^{8} (x - 1)^{7})

\int (2 x - 4) d x

\int (- \frac{3}{7} x^{5} + \frac{1}{5} x^{3} - \frac{7}{4} x^{2}) d x