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

Lösung

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

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

(d^{2} - 8 d + 9) y = 8 sin^{5} (x)

(d^{2} + 5 d + 6) \cdot y = x^{2}

(d^{3} + 8) y = x^{4} + 2 x + 1

d \cdot e (d^{2} - 4) y = (e^{2})^{x}

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