What is the solution for y^{''}+3y=1 ?

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

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y^{''}+3y=1

y^{^{''}} + 3 y = 1

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

Solution steps

y^{''} (t) + 3 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}

(x y - y) d y - (y + 1) d x = 0

\int sin (x) cos^{2} (x) + sin^{3} (x) d x

\frac{\partial}{\partial y} (\frac{e ^{2 - x}}{y ^{2}})

\frac{d}{d x} (\frac{2 + x ^{3}}{2 - x ^{3}})

f (x) = tan (x) - sin (x)

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