What is the solution for 5x^{''}+50x-10=0 ?

The solution for 5x^{''}+50x-10=0 is x=c_{1}cos(sqrt(10)t)+c_{2}sin(sqrt(10)t)+1/5

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5x^{''}+50x-10=0

5 x^{^{''}} + 50 x - 10 = 0

x = c_{1} cos (10 t) + c_{2} sin (10 t) + \frac{1}{5}

Solution steps

5 x^{''} (t) + 50 x - 10 = 0

Solve linear ODE:

x = c_{1} cos (10 t) + c_{2} sin (10 t) + \frac{1}{5}

x = c_{1} cos (10 t) + c_{2} sin (10 t) + \frac{1}{5}

\frac{\partial}{\partial y} ((2 x + 3 y)^{7})

\frac{d}{d u} (ln (u cos (v)))

\frac{d y}{d x} = x y, y (0) = 2

f (x) = 2 e^{2 x - 4} - e^{2 x - 4}

\int_{0}^{5} 2 π y (25 - y^{2}) d y

What is the solution for 5x^{''}+50x-10=0 ?
The solution for 5x^{''}+50x-10=0 is x=c_{1}cos(sqrt(10)t)+c_{2}sin(sqrt(10)t)+1/5