What is the solution for y^{''}+4y=2,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+4y=2,y(0)=0,y^'(0)=0 is y=-1/2 cos(2t)+1/2

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

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

y = - \frac{1}{2} cos (2 t) + \frac{1}{2}

Solution steps

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

Solve linear ODE:

y = - \frac{1}{2} cos (2 t) + \frac{1}{2}

y = - \frac{1}{2} cos (2 t) + \frac{1}{2}

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

y^{^{''}} - y^{^{'}} + \frac{1}{4} y = x e^{\frac{x}{2}}

4 x^{^{''}} - 2 x = 7

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

y^{^{''}} + 4 y^{^{'}} + 4 = \frac{e ^{- 2 x}}{x ^{3}}

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