(dy)/(dt)+y=te^t

Lösung

\frac{d y}{d t} + y = t e^{t}

y = \frac{e ^{t} ( 2 t - 1 )}{4} + \frac{c _{1}}{e ^{t}}

Schritte zur Lösung

\frac{d y}{d t} + y = t e^{t}

Solve linear ODE:

y = \frac{e ^{t} ( 2 t - 1 )}{4} + \frac{c _{1}}{e ^{t}}

y = \frac{e ^{t} ( 2 t - 1 )}{4} + \frac{c _{1}}{e ^{t}}

\int_{- 1}^{1} 3 x d x

t an g e n t f (x) = \frac{1}{2 x}, at x = 36

d er i v a t i v e f (x) = \frac{tan ( 9 x )}{1 + sin ( 9 x )}

(2 x + 3 y - 1) d x - 4 (x + 1) d y = 0

in v erse l a pl a ce \frac{e ^{- 2 s}}{s ( s - 2 )}