What is the solution for y^{''}+2y=10 ?

The solution for y^{''}+2y=10 is y=c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t)+5

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y^{''}+2y=10

y^{^{''}} + 2 y = 10

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 5

Solution steps

y^{''} (t) + 2 y = 10

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 5

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 5

\frac{( s + 1 )}{s ^{2} + s + 1}

\frac{d y}{d x} = x e^{- y}

\int_{7}^{15} ln (x^{2}) d x

y^{^{'}} = \frac{x ^{8} - 2 y}{x}

ln (csc (x) - cot (x))

What is the solution for y^{''}+2y=10 ?
The solution for y^{''}+2y=10 is y=c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t)+5