t^2y^{''}+ty^'+4y=sin(2ln(t))

Lösung

t^{2} y^{^{''}} + t y^{^{'}} + 4 y = sin (2 ln (t))

y = c_{1} cos (2 ln (t)) + c_{2} sin (2 ln (t)) + \frac{- cos ( 2 ln ( t ) ) ( 4 ln ( t ) - sin ( 4 ln ( t ) ) ) - cos ( 4 ln ( t ) ) sin ( 2 ln ( t ) )}{16}

Schritte zur Lösung

t^{2} y^{''} (t) + t y^{'} (t) + 4 y = sin (2 ln (t))

Solve linear ODE:

y = c_{1} cos (2 ln (t)) + c_{2} sin (2 ln (t)) + \frac{- cos ( 2 ln ( t ) ) ( 4 ln ( t ) - sin ( 4 ln ( t ) ) ) - cos ( 4 ln ( t ) ) sin ( 2 ln ( t ) )}{16}

y = c_{1} cos (2 ln (t)) + c_{2} sin (2 ln (t)) + \frac{- cos ( 2 ln ( t ) ) ( 4 ln ( t ) - sin ( 4 ln ( t ) ) ) - cos ( 4 ln ( t ) ) sin ( 2 ln ( t ) )}{16}

\int - 5 tan^{5} (4 x) sec^{6} (4 x) d x

x \to 0 lim (\frac{tan ( 9 x )}{sin ( 4 x )})

\int \frac{x}{4 x + 4} d x

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

s l o p e in t erce pt (1, 1), (3, - \frac{5}{6})