y^{''}+9y=2sin(2x),y(0)=0,y^'(0)=-1

解答

y^{^{''}} + 9 y = 2 sin (2 x), y (0) = 0, y^{^{'}} (0) = - 1

y = - \frac{3}{5} sin (3 x) + \frac{2}{5} sin (2 x)

求解步骤

y^{''} (x) + 9 y = 2 sin (2 x)

求解线性方程:

y = - \frac{3}{5} sin (3 x) + \frac{2}{5} sin (2 x)

y = - \frac{3}{5} sin (3 x) + \frac{2}{5} sin (2 x)

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