ty^'+4y=0

Lösung

t y^{^{'}} + 4 y = 0

y = \frac{c _{1}}{t ^{4}}

Schritte zur Lösung

t y^{'} (t) + 4 y = 0

Solve separable ODE:

y = \frac{c _{1}}{t ^{4}}

y = \frac{c _{1}}{t ^{4}}

\int \frac{1}{y ^{2} + 3} d y

n = 0 \sum \infty n (z - 1)^{n - 1}

\frac{\partial}{\partial t} (6 q (r, t)^{2} + 7 r^{3} - \frac{8}{t} - q (r, t) t)

n = 0 \sum \infty \frac{2}{n ^{2} + 6 n + 8}

d er i v a t i v e (2 x - 5)^{2}