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Popular >

y^{''}+4y^'+29y=2,y^'(0)=3

Solution

y^{^{''}} + 4 y^{^{'}} + 29 y = 2, y^{^{'}} (0) = 3

Solution

y = e^{- 2 t} (- \frac{cos ( 5 t ) ( 3 - 5 c _{2} )}{2} + c_{2} sin (5 t)) + \frac{2}{29}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 29 y = 2

Solve linear ODE:

y = e^{- 2 t} (- \frac{cos ( 5 t ) ( 3 - 5 c _{2} )}{2} + c_{2} sin (5 t)) + \frac{2}{29}

y = e^{- 2 t} (- \frac{cos ( 5 t ) ( 3 - 5 c _{2} )}{2} + c_{2} sin (5 t)) + \frac{2}{29}

Graph

Popular Examples

y^{''}-5y^'+6y=4e^{4x}

y^{^{''}} - 5 y^{^{'}} + 6 y = 4 e^{4 x}

y^{''}+2y^'+3y=5sin(4t),y(0)=2,y^'(0)=1

y^{^{''}} + 2 y^{^{'}} + 3 y = 5 sin (4 t), y (0) = 2, y^{^{'}} (0) = 1

y^{''}-3y^'-4y=-16t,y(0)=-4,y^'(0)=-5

y^{^{''}} - 3 y^{^{'}} - 4 y = - 16 t, y (0) = - 4, y^{^{'}} (0) = - 5

y^{''}+4y=5sin(2x)-7cos(2x)

y^{^{''}} + 4 y = 5 sin (2 x) - 7 cos (2 x)

y^{''}-4y^'=6e^{3x}-3e^{-x}

y^{^{''}} - 4 y^{^{'}} = 6 e^{3 x} - 3 e^{- x}

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+4y^'+29y=2,y^'(0)=3 ?
The solution for y^{''}+4y^'+29y=2,y^'(0)=3 is y=e^{-2t}(-(cos(5t)(3-5c_{2}))/2+c_{2}sin(5t))+2/29