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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

y^{''}+2y^'+y=sin(t)

y^{^{''}} + 2 y^{^{'}} + y = sin (t)

y^{''}-8y^'+16y=12e^{4x}

y^{^{''}} - 8 y^{^{'}} + 16 y = 12 e^{4 x}

y^{''}+2y^'-8y=3e^{-3x}-e^{-x}

y^{^{''}} + 2 y^{^{'}} - 8 y = 3 e^{- 3 x} - e^{- x}

x^{''}+1.1875x^'+0.375=0

x^{^{''}} + 1.1875 x^{^{'}} + 0.375 = 0

y^{''}-y^'-2y=4,x^2y(0)=1,y^'(0)=4

y^{^{''}} - y^{^{'}} - 2 y = 4, x^{2} y (0) = 1, y^{^{'}} (0) = 4

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