y^'=8x^3+2 y/x

Lösung

y^{^{'}} = 8 x^{3} + 2 \frac{y}{x}

y = 4 x^{4} + c_{1} x^{2}

Schritte zur Lösung

y^{'} (x) = 8 x^{3} + 2 \frac{y}{x}

Solve linear ODE:

y = 4 x^{4} + c_{1} x^{2}

y = 4 x^{4} + c_{1} x^{2}

\frac{d y}{d t} = - 2 y, y (0) = 2

n = 1 \sum \infty n (\frac{5}{4})^{n}

\frac{\partial}{\partial x} (x^{3} y^{5})

n = 2 \sum \infty \frac{1}{n - 1} - \frac{1}{n}

\int \frac{x ^{2} + 3 x - 1}{x ^{3} + 8} d x