x^{''}+4x=-t^2

Lösung

x^{^{''}} + 4 x = - t^{2}

x = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{t ^{2}}{4} + \frac{1}{8}

Schritte zur Lösung

x^{''} (t) + 4 x = - t^{2}

Solve linear ODE:

x = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{t ^{2}}{4} + \frac{1}{8}

x = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{t ^{2}}{4} + \frac{1}{8}

y^{^{''}} + y = 5 sec^{3} (x)

4 y^{^{''}} - 16 y = 4 x

x^{^{''}} + 2 x = 2 t + 26

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

y^{^{''}} - 14 y^{^{'}} + 49 y = e^{7 t} ln (t)