y^{''}+4y=(sin(x))^2

Lösung

y^{^{''}} + 4 y = (sin (x))^{2}

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

Schritte zur Lösung

y^{''} (x) + 4 y = (sin (x))^{2}

Solve linear ODE:

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

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

x^{^{''}} + x^{^{'}} + 4 = 0

y^{^{''}} + 4 y = (sin (x))^{3}

4 y^{^{''}} + 5 y^{^{'}} + 2 y = - 2 x^{2} + x

y^{^{''}} - y^{^{'}} + y = 2 cos (2 x) - 3 sin (2 x)

y^{^{'''}} + 4 y^{^{''}} - 9 y^{^{'}} - 36 = x^{2} - 4