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

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

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

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

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

Solution steps

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

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}

\int x (5 x^{2} + \frac{2}{3})^{4} d x

\frac{d}{d x} (- 5 x^{3} + x^{2} - 3 x - 3)

\int_{0}^{1} \frac{1}{1 + x ^{2}} d x

\int \frac{2 x + 7}{x ^{2} + 2 x - 4} d x

\frac{d}{d x} (\frac{cos ( x )}{1 + co tan ( x )})

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