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

Lösung

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{5} sin^{5} (x) cos (2 x) + sin (2 x) \cdot \int \frac{sin ^{3} ( x ) cos ( 2 x )}{2} d x

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{5} sin^{5} (x) cos (2 x) + sin (2 x) \cdot \int \frac{sin ^{3} ( x ) cos ( 2 x )}{2} d x

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{5} sin^{5} (x) cos (2 x) + sin (2 x) \cdot \int \frac{sin ^{3} ( x ) cos ( 2 x )}{2} d x

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

y^{^{''}} + y = \frac{1}{( cos ( x ) )}

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