y^'=t^2-y

Lösung

y^{^{'}} = t^{2} - y

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

Schritte zur Lösung

y^{'} (t) = t^{2} - y

Solve linear ODE:

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

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

\frac{d}{d x} (arccos (\frac{3}{x}))

7 y^{^{''}} + 4 y^{^{'}} = 0

\int (t^{2} + 2)^{3} d t

\frac{d}{d t} (6 sin (t))

t an g e n t y = 4 x^{2}