(d^2+4)y=tan^2(x)

Lösung

(d^{2} + 4) y = tan^{2} (x)

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

Schritte zur Lösung

(d^{2} + 4) y = tan^{2} (x)

Solve linear ODE:

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

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

(d^{3} - 1) \cdot y = e^{x}

y^{^{''}} + 4 y = 0, y^{1} (0) = sin^{2} (x)

2 y^{^{''}} - 72 y = 0

y^{^{''}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 5

u^{^{''}} \frac{( x )}{x} + b^{2} H (x) \frac{( x )}{x} = 0