y^{''}+4y=3x(2+cos(2x))

Lösung

y^{^{''}} + 4 y = 3 x (2 + cos (2 x))

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{3 x}{2} + \frac{3 x ^{2} sin ( 2 x )}{8} + \frac{3 x cos ( 2 x )}{16}

Schritte zur Lösung

y^{''} (x) + 4 y = 3 x (2 + cos (2 x))

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{3 x}{2} + \frac{3 x ^{2} sin ( 2 x )}{8} + \frac{3 x cos ( 2 x )}{16}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{3 x}{2} + \frac{3 x ^{2} sin ( 2 x )}{8} + \frac{3 x cos ( 2 x )}{16}

l in e a r y^{^{''}} - 2 y^{^{'}} + 5 y = 6 e^{t}, y (0) = 0

y^{^{''}} + 3 y^{^{'}} + 2 y = 10 u, y (0) = 0, y^{^{'}} (0) = 0

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

\frac{d ^{2} y}{d x ^{2}} - y - 6 = 1

y^{^{''}} + y^{^{'}} + 2 y = x + 3, y (0) = 0, y^{^{'}} (0) = 1