y^{''}-2y^'+5y=e^tcos(2t)

Lösung

y^{^{''}} - 2 y^{^{'}} + 5 y = e^{t} cos (2 t)

y = e^{t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{e ^{t} t sin ( 2 t )}{4}

Schritte zur Lösung

y^{''} (t) - 2 y^{'} (t) + 5 y = e^{t} cos (2 t)

Solve linear ODE:

y = e^{t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{e ^{t} t sin ( 2 t )}{4}

y = e^{t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{e ^{t} t sin ( 2 t )}{4}

l a pl a ce t r an s f or m (sin (6 t) - 7)^{2}

\int \frac{cos ^{2} ( θ )}{sin ( θ )} d θ

d er i v a t i v e f (x) = 2^{90}

x \to 5 + lim (x^{3} - c x)

2 y^{^{''}} - 4 y^{^{'}} + 3 y = 9 t^{2} + 10