ty^'+(t+1)y=2te^{-t}

Lösung

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

y = \frac{t}{e ^{t}} + \frac{c _{1}}{e ^{t} t}

Schritte zur Lösung

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

Solve linear ODE:

y = \frac{t}{e ^{t}} + \frac{c _{1}}{e ^{t} t}

y = \frac{t}{e ^{t}} + \frac{c _{1}}{e ^{t} t}

in v erse l a pl a ce \frac{2 s + 12}{s ^{2} + 6 s + 18}

\int \frac{5 x ^{3} - 4 x}{1 - x ^{4}} d x

\int_{0}^{1} \frac{x - 7}{x ^{2} - 5 x + 6} d x

y^{^{'}} + 6 y = e^{4 t}

\frac{d}{d x} (sec^{5} (x^{2} - 1))