y^{''}+4y=-6,y(pi/8)=-1/2 ,y^'(pi/8)=2

解答

y^{^{''}} + 4 y = - 6, y (\frac{π}{8}) = - \frac{1}{2}, y^{^{'}} (\frac{π}{8}) = 2

y = 2 sin (2 t) - \frac{3}{2}

求解步骤

y^{''} (t) + 4 y = - 6

求解线性方程:

y = 2 sin (2 t) - \frac{3}{2}

y = 2 sin (2 t) - \frac{3}{2}

y^{^{''}} - 4 y^{^{'}} + 8 y = 8 x^{2} - 2

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

(D^{2} + 1) y = sec (x) tan (x)

0.5 q^{^{''}} + 6 q^{^{'}} + \frac{1}{0.02} q = 24 sin (10 t)

y^{^{''}} - 3 y^{^{'}} + 2 y = 13 e^{3 x}