What is the solution for y^{''}+2y+y=1 ?

The solution for y^{''}+2y+y=1 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)+1/3

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y^{''}+2y+y=1

y^{^{''}} + 2 y + y = 1

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{3}

Solution steps

y^{''} (t) + 2 y + y = 1

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{3}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{3}

\frac{\partial}{\partial x} (\frac{x}{( x ^{2} + y ^{2} ) ^{2}})

\int x^{2} 2 - x^{3} d x

cos (x), at \frac{π}{2}

\frac{\partial}{\partial y} ((x - y) (1 - x y))

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

What is the solution for y^{''}+2y+y=1 ?
The solution for y^{''}+2y+y=1 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)+1/3