y^'+y=1

Lösung

y^{^{'}} + y = 1

y = - e^{- t - c_{1}} + 1

Schritte zur Lösung

y^{'} (t) + y = 1

Solve separable ODE:

y = - e^{- t - c_{1}} + 1

y = - e^{- t - c_{1}} + 1

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

\int \frac{x}{x ^{3}} d x

\int (x^{3} + e^{x}) d x

n = 1 \sum \infty n^{0.5}

t an g e n t \frac{2 x}{x + 1}