y^{''}+0.2121y^'+2y=cos(4x),y(0)=0

Lösung

y^{^{''}} + 0.2121 y^{^{'}} + 2 y = cos (4 x), y (0) = 0

y = e^{- 0.10605 x} (0.07116 \dots cos (1.41023 \dots x) + c_{2} sin (1.41023 \dots x)) + 0.00431 \dots sin (4 x) - 0.07116 \dots cos (4 x)

Schritte zur Lösung

y^{''} (x) + 0.2121 y^{'} (x) + 2 y = cos (4 x)

Solve linear ODE:

y = e^{- 0.10605 x} (0.07116 \dots cos (1.41023 \dots x) + c_{2} sin (1.41023 \dots x)) + 0.00431 \dots sin (4 x) - 0.07116 \dots cos (4 x)

y = e^{- 0.10605 x} (0.07116 \dots cos (1.41023 \dots x) + c_{2} sin (1.41023 \dots x)) + 0.00431 \dots sin (4 x) - 0.07116 \dots cos (4 x)

\int \frac{( ln ( x ^{15} ) )}{x} d x

t an g e n t 5 x e^{x}

\int_{1}^{7} \frac{5}{x} d x

a re a x = y^{2} + 6 y, y = x

n = 0 \sum \infty x^{n + 5}