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Popular >

y^{''}+4y+4=0

Solution

y^{^{''}} + 4 y + 4 = 0

Solution

y = c_{1} cos (2 t) + c_{2} sin (2 t) - 1

Solution steps

y^{''} (t) + 4 y + 4 = 0

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) - 1

y = c_{1} cos (2 t) + c_{2} sin (2 t) - 1

Graph

Popular Examples

x^{''}+8x^'+16x=e^{-t}sin(4t)

x^{^{''}} + 8 x^{^{'}} + 16 x = e^{- t} sin (4 t)

D(t)=(100ln(24+2))/(24+2)

D (t) = \frac{100 ln ( 24 + 2 )}{24 + 2}

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

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

y^{^{''}} + y^{^{'}} = 4 t

y^{^{''}} - y + t^{2} = 0

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+4y+4=0 ?
The solution for y^{''}+4y+4=0 is y=c_{1}cos(2t)+c_{2}sin(2t)-1