y^{''}+6y=4t^2-3,y(0)=0,y^'(0)=-7

Lösung

y^{^{''}} + 6 y = 4 t^{2} - 3, y (0) = 0, y^{^{'}} (0) = - 7

y = \frac{13}{18} cos (6 t) - \frac{7}{6} sin (6 t) + \frac{2 t ^{2}}{3} - \frac{13}{18}

Schritte zur Lösung

y^{''} (t) + 6 y = 4 t^{2} - 3

Solve linear ODE:

y = \frac{13}{18} cos (6 t) - \frac{7}{6} sin (6 t) + \frac{2 t ^{2}}{3} - \frac{13}{18}

y = \frac{13}{18} cos (6 t) - \frac{7}{6} sin (6 t) + \frac{2 t ^{2}}{3} - \frac{13}{18}

y^{^{''}} + 10 y^{^{'}} + 16 y = 2581 cos (5 x)

y^{^{''}} - 3 y^{^{'}} + 2 y = \frac{e ^{x}}{1 + e ^{x}}

y^{^{''}} + y^{^{'}} + \frac{1}{4} y = e^{x} (sin (3 x) - cos (3 x))

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x}}{( 1 + x ^{2} )}

y^{^{''}} + 3 y^{^{'}} + 2 y = 6 e^{x}