y^{''}+2y= 1/(cos(sqrt(2x)))

Lösung

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

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

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

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

y^{^{''}} - 3 y = - 48 x^{2} e^{3 x}

y^{^{''}} + 16 y = 3 cos (4 t)

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