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

解答

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

y = - \frac{cos ( 3 t ) ( 1 - 3 cos ( 3 ) c _{2} )}{3 sin ( 3 )} + c_{2} sin (3 t)

求解步骤

y^{''} (t) + 3 y = 0

求解线性方程:

y = - \frac{cos ( 3 t ) ( 1 - 3 cos ( 3 ) c _{2} )}{3 sin ( 3 )} + c_{2} sin (3 t)

y = - \frac{cos ( 3 t ) ( 1 - 3 cos ( 3 ) c _{2} )}{3 sin ( 3 )} + c_{2} sin (3 t)

16 y^{^{''}} + 40 y^{^{'}} + 50 y = 0

y^{^{''}} - 16 y = 0, y (0) = e^{4 x}, y (0) = 9, y^{^{'}} (0) = 3

y^{^{''}} + y^{^{'}} + 24 y = 0, y (0) = 0, y^{^{'}} (0) = - 7

0.001 y^{^{''}} + 10 y = 0

y^{^{''}} + 1 y - 2 y^{^{'}} = 0