(d^3+4d)y=sin^2(x)

Lösung

(d^{3} + 4 d) y = sin^{2} (x)

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{1}{8} (x - \frac{1}{2} sin (2 x)) - \frac{1}{4} cos (2 x) (\frac{1}{2} sin^{2} (x) sin (2 x) - \frac{1}{4} (x - \frac{1}{4} sin (4 x))) - \frac{1}{8} sin^{4} (x) sin (2 x)

Schritte zur Lösung

(d^{3} + 4 d) y = sin^{2} (x)

Solve linear ODE:

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{1}{8} (x - \frac{1}{2} sin (2 x)) - \frac{1}{4} cos (2 x) (\frac{1}{2} sin^{2} (x) sin (2 x) - \frac{1}{4} (x - \frac{1}{4} sin (4 x))) - \frac{1}{8} sin^{4} (x) sin (2 x)

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{1}{8} (x - \frac{1}{2} sin (2 x)) - \frac{1}{4} cos (2 x) (\frac{1}{2} sin^{2} (x) sin (2 x) - \frac{1}{4} (x - \frac{1}{4} sin (4 x))) - \frac{1}{8} sin^{4} (x) sin (2 x)

y^{^{''}} + y^{^{'}} = t^{2} + 2 t

y^{^{''}} + 9 y = 10, e^{- t} y (0) y^{^{'}} (0) = 0

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

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

y^{^{''}} + y^{^{'}} = 1, y^{1} (0) = 1