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Popular >

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

Solution

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

Solution

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

Solution steps

2 y^{''} (t) + 2 y = 0

Solve linear ODE:

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

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

Graph

Popular Examples

(D^3+D^2+4D+4)y=0

(D^{3} + D^{2} + 4 D + 4) y = 0

(d^4y)/(dx^4)+y=0

\frac{d ^{4} y}{d x ^{4}} + y = 0

y^{'''}-7y^{''}+19y^'-13y=0

y^{^{'''}} - 7 y^{^{''}} + 19 y^{^{'}} - 13 y = 0

y^{'''}-3y^{''}-9y^'+27y=0

y^{^{'''}} - 3 y^{^{''}} - 9 y^{^{'}} + 27 y = 0

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

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

Frequently Asked Questions (FAQ)

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