y^{''}-2y^'+10y=e^xcos(3x)

Lösung

y^{^{''}} - 2 y^{^{'}} + 10 y = e^{x} cos (3 x)

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{e ^{x} x sin ( 3 x )}{6}

Schritte zur Lösung

y^{''} (x) - 2 y^{'} (x) + 10 y = e^{x} cos (3 x)

Solve linear ODE:

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{e ^{x} x sin ( 3 x )}{6}

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{e ^{x} x sin ( 3 x )}{6}

\int sin (u^{2}) d u

\frac{\partial}{\partial y} (\frac{2 y}{x ^{2} + y ^{2}})

d er i v a t i v e \frac{x ^{2}}{8 + x}

\int \frac{cos ( x )}{sin ( x )} d x

n = 0 \sum \infty \frac{1}{n} - \frac{1}{n + 2}