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Popular >

x^{''}+6x^'+25x=0,x(0)= 1/2 ,x^'(0)=0

Solution

x^{^{''}} + 6 x^{^{'}} + 25 x = 0, x (0) = \frac{1}{2}, x^{^{'}} (0) = 0

Solution

x = e^{- 3 t} (\frac{1}{2} cos (4 t) + \frac{3}{8} sin (4 t))

Solution steps

x^{''} (t) + 6 x^{'} (t) + 25 x = 0

Solve linear ODE:

x = e^{- 3 t} (\frac{1}{2} cos (4 t) + \frac{3}{8} sin (4 t))

x = e^{- 3 t} (\frac{1}{2} cos (4 t) + \frac{3}{8} sin (4 t))

Popular Examples

4y^{''}-12y^'+9y=0,y^1(0)=e^{(3x)/2}

4 y^{^{''}} - 12 y^{^{'}} + 9 y = 0, y^{1} (0) = e^{\frac{3 x}{2}}

81y^{''}-90y^'+106y=0

81 y^{^{''}} - 90 y^{^{'}} + 106 y = 0

y^{''}+4y^'+13y=0,y(0)=4,y^'(0)=-7

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 4, y^{^{'}} (0) = - 7

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

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

y^{''}+14y^'+51y=0

y^{^{''}} + 14 y^{^{'}} + 51 y = 0

Frequently Asked Questions (FAQ)

What is the solution for x^{''}+6x^'+25x=0,x(0)= 1/2 ,x^'(0)=0 ?
The solution for x^{''}+6x^'+25x=0,x(0)= 1/2 ,x^'(0)=0 is x=e^{-3t}(1/2 cos(4t)+3/8 sin(4t))