y^{''}+y=sin(x),y^'(pi/2)= 3/2 ,y(0)=0

解答

y^{^{''}} + y = sin (x), y^{^{'}} (\frac{π}{2}) = \frac{3}{2}, y (0) = 0

y = - \frac{6 - π}{4} cos (x) + c_{2} sin (x) - \frac{x cos ( x )}{2}

求解步骤

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

求解线性方程:

y = - \frac{6 - π}{4} cos (x) + c_{2} sin (x) - \frac{x cos ( x )}{2}

y = - \frac{6 - π}{4} cos (x) + c_{2} sin (x) - \frac{x cos ( x )}{2}

\frac{d ^{3} y}{d x ^{3}} + \frac{d y}{d x} = tan (x)

3 x^{^{''}} + 25 x^{^{'}} + 48 x = 5 cos (4 t)

y^{^{''}} + 4 y = t^{2} + 6 e^{t}, y (0) = 0, y^{^{'}} (0) = 5

y^{^{''}} - 8 y^{^{'}} + 20 y = 100 x^{2} - 39 x e^{x}

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