What is the solution for x^{''}+2x^'+10x=0,x(0)=1,x^'(0)=0 ?

The solution for x^{''}+2x^'+10x=0,x(0)=1,x^'(0)=0 is x=e^{-t}(cos(3t)+1/3 sin(3t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}+2x^'+10x=0,x(0)=1,x^'(0)=0

x^{^{''}} + 2 x^{^{'}} + 10 x = 0, x (0) = 1, x^{^{'}} (0) = 0

x = e^{- t} (cos (3 t) + \frac{1}{3} sin (3 t))

Solution steps

x^{''} (t) + 2 x^{'} (t) + 10 x = 0

Solve linear ODE:

x = e^{- t} (cos (3 t) + \frac{1}{3} sin (3 t))

x = e^{- t} (cos (3 t) + \frac{1}{3} sin (3 t))

4 y^{^{''}} + \frac{16}{0.25} y = 0, y (0) = 0, y^{^{'}} (0) = 0

\frac{16}{32} x^{^{''}} + x^{^{'}} + 2 x = 0, x (0) = - 2, x^{^{'}} (0) = 1

100 y^{^{'''}} + 60 y^{^{''}} + 9 y^{^{'}} = 0

8 y - 2 y^{^{'}} + 6 y^{^{''}} - y^{^{'}} = 0

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y^{^{'}} 0 = (3), y (0) = 0

What is the solution for x^{''}+2x^'+10x=0,x(0)=1,x^'(0)=0 ?
The solution for x^{''}+2x^'+10x=0,x(0)=1,x^'(0)=0 is x=e^{-t}(cos(3t)+1/3 sin(3t))