y^{''}-2y^'+5y=1+x,y(0)=0,y^'(0)=4

Lösung

y^{^{''}} - 2 y^{^{'}} + 5 y = 1 + x, y (0) = 0, y^{^{'}} (0) = 4

y = e^{x} (- \frac{7}{25} cos (2 x) + \frac{51}{25} sin (2 x)) + \frac{x}{5} + \frac{7}{25}

Schritte zur Lösung

y^{''} (x) - 2 y^{'} (x) + 5 y = 1 + x

Solve linear ODE:

y = e^{x} (- \frac{7}{25} cos (2 x) + \frac{51}{25} sin (2 x)) + \frac{x}{5} + \frac{7}{25}

y = e^{x} (- \frac{7}{25} cos (2 x) + \frac{51}{25} sin (2 x)) + \frac{x}{5} + \frac{7}{25}

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

y^{^{''}} + 2 y - 1 = 0

y^{^{''}} + 5 y^{^{'}} + 6 y = 6 t + 17, y (0) = 3, y^{^{'}} (0) = - 4

y^{^{''}} - 3 y^{^{'}} + 2 y = t - 2, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 6 y^{^{'}} + 5 y = e^{2 t} \cdot 3