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

Lösung

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

x = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}) + t^{2} - 8 - \frac{2}{3} cos (t)

Schritte zur Lösung

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

Solve linear ODE:

x = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}) + t^{2} - 8 - \frac{2}{3} cos (t)

x = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}) + t^{2} - 8 - \frac{2}{3} cos (t)

y^{^{''}} - 4 y + 4 = - 5 e^{3 x}

x^{^{''}} + 4 x = cos (4 t)

x^{^{''}} - 2 x^{^{'}} + 3 x + 2 = 0

y^{^{''}} - 10 y^{^{'}} + 16 = 0

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