y^{''}+2y=sin(sqrt(2)t),y(0)=0,y^'(0)=0

Lösung

y^{^{''}} + 2 y = sin (2 t), y (0) = 0, y^{^{'}} (0) = 0

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

Schritte zur Lösung

y^{''} (t) + 2 y = sin (2 t)

Solve linear ODE:

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

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

\int \frac{1}{sec ( x ) + tan ( x )} d x

in v erse l a pl a ce \frac{1}{s ^{3} + 8 s ^{2} + 12 s}

d er i v a t i v e f (x) = 4 x^{\frac{4}{3}} - \frac{2}{3} x^{\frac{3}{2}} + x^{2} - 4 x + 1

n = 0 \sum \infty 2 \cdot 0. 2^{n}

x \to 0 lim (\frac{5 ^{x + 1} - 5}{x})