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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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}

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