y^{''}+9y=e^t,y(0)=0,y^'(0)=0

Lösung

y^{^{''}} + 9 y = e^{t}, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{1}{10} cos (3 t) - \frac{1}{30} sin (3 t) + \frac{1}{10} e^{t}

Schritte zur Lösung

y^{''} (t) + 9 y = e^{t}

Solve linear ODE:

y = - \frac{1}{10} cos (3 t) - \frac{1}{30} sin (3 t) + \frac{1}{10} e^{t}

y = - \frac{1}{10} cos (3 t) - \frac{1}{30} sin (3 t) + \frac{1}{10} e^{t}

\int e^{(3 x + 1)} d x

x \to 27 lim (\frac{x - 27}{x + 9 - 6})

a re a 4 x, x^{2}

\int_{3}^{9} \frac{1}{x ln ( x )} d x

t an g e n t y = 2 x + 1, at x = \frac{1}{2}