y^{''}-4y^'+5y=e^{2x}

Lösung

y^{^{''}} - 4 y^{^{'}} + 5 y = e^{2 x}

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) + e^{2 x}

Schritte zur Lösung

y^{''} (x) - 4 y^{'} (x) + 5 y = e^{2 x}

Solve linear ODE:

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) + e^{2 x}

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) + e^{2 x}

\frac{\partial}{\partial x} ((1 + x y) e^{x y})

y^{^{'}} = 1 - x + y - x y

n = 1 \sum \infty \frac{4 ^{n}}{n + 1}

d er i v a t i v e f (x) = 4 x^{4} - 3 x^{\frac{5}{2}} + 2

\frac{\partial}{\partial x} (3 e^{y} + x + x^{3} + 6)