ty^'-2y=t^5sin(2t)-t^3+4t^4

Lösung

t y^{^{'}} - 2 y = t^{5} sin (2 t) - t^{3} + 4 t^{4}

y = - \frac{1}{2} t^{4} cos (2 t) + \frac{1}{4} t^{2} (2 t sin (2 t) + cos (2 t)) - t^{3} + 2 t^{4} + c_{1} t^{2}

Schritte zur Lösung

t y^{'} (t) - 2 y = t^{5} sin (2 t) - t^{3} + 4 t^{4}

Solve linear ODE:

y = - \frac{1}{2} t^{4} cos (2 t) + \frac{1}{4} t^{2} (2 t sin (2 t) + cos (2 t)) - t^{3} + 2 t^{4} + c_{1} t^{2}

y = - \frac{1}{2} t^{4} cos (2 t) + \frac{1}{4} t^{2} (2 t sin (2 t) + cos (2 t)) - t^{3} + 2 t^{4} + c_{1} t^{2}

t an g e n t f (x) = \frac{12 x}{x ^{2} - 4}, at x = 4

h \to 0 lim (\frac{- 8 + h}{h})

(e^{- \frac{1}{x ^{2}}})^{^{'}}

\int \frac{e ^{x}}{e ^{2 x} + 1} d x

\int_{2}^{3} \frac{5}{3 - x} d x