4y^{''}-4y^'+y=e^{x/2}sqrt(1-x^2)

Lösung

4 y^{^{''}} - 4 y^{^{'}} + y = e^{\frac{x}{2}} 1 - x^{2}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + \frac{1}{12} (1 - x^{2})^{\frac{3}{2}} e^{\frac{x}{2}} + \frac{1}{8} (arcsin (x) + \frac{1}{2} sin (2 arcsin (x))) e^{\frac{x}{2}} x

Schritte zur Lösung

4 y^{''} (x) - 4 y^{'} (x) + y = e^{\frac{x}{2}} 1 - x^{2}

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + \frac{1}{12} (1 - x^{2})^{\frac{3}{2}} e^{\frac{x}{2}} + \frac{1}{8} (arcsin (x) + \frac{1}{2} sin (2 arcsin (x))) e^{\frac{x}{2}} x

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + \frac{1}{12} (1 - x^{2})^{\frac{3}{2}} e^{\frac{x}{2}} + \frac{1}{8} (arcsin (x) + \frac{1}{2} sin (2 arcsin (x))) e^{\frac{x}{2}} x

y^{^{'''}} + y^{^{'}} = sec (t)

y^{^{''}} - 4 y = \frac{e ^{2 x}}{x}

y^{^{'''}} + y^{^{'}} = tan (t)

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

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