(d^2+9)y=sec(3x)

Lösung

(d^{2} + 9) y = sec (3 x)

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{9} ln (cos (3 x)) cos (3 x) + \frac{1}{3} x sin (3 x)

Schritte zur Lösung

(d^{2} + 9) y = sec (3 x)

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{9} ln (cos (3 x)) cos (3 x) + \frac{1}{3} x sin (3 x)

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{9} ln (cos (3 x)) cos (3 x) + \frac{1}{3} x sin (3 x)

(d^{2} - 3 d + 2) \cdot y = e^{5 x}

y^{^{''}} - y = \frac{1}{x}

y^{^{'''}} - y^{^{'}} = 4 e^{- x} + 3 e^{2 x}

(d^{2} - 4 d + 3) y = x^{3}

(d + 1) (d - 1)^{2} y = e^{- x}