What is the solution for 10y^{''}+6y^'+y=0 ?

The solution for 10y^{''}+6y^'+y=0 is y=e^{-(3t)/(10)}(c_{1}cos(t/(10))+c_{2}sin(t/(10)))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

10y^{''}+6y^'+y=0

10 y^{^{''}} + 6 y^{^{'}} + y = 0

y = e^{- \frac{3 t}{10}} (c_{1} cos (\frac{t}{10}) + c_{2} sin (\frac{t}{10}))

Solution steps

10 y^{''} (t) + 6 y^{'} (t) + y = 0

Solve linear ODE:

y = e^{- \frac{3 t}{10}} (c_{1} cos (\frac{t}{10}) + c_{2} sin (\frac{t}{10}))

y = e^{- \frac{3 t}{10}} (c_{1} cos (\frac{t}{10}) + c_{2} sin (\frac{t}{10}))

u^{^{''}} + 14 H (t) = 0

9 y^{^{''}} + 30 y^{^{'}} - 25 y = 0

\frac{d ^{2} y}{d t ^{2}} - 4 \frac{d y}{d t} - 5 y = 0, y (1) = 0

x^{^{''}} = - a^{2} x

y^{^{''}} + 4 y^{^{'}} + 29 y = 0, y (0) = 2, y^{^{'}} (0) = - 3

What is the solution for 10y^{''}+6y^'+y=0 ?
The solution for 10y^{''}+6y^'+y=0 is y=e^{-(3t)/(10)}(c_{1}cos(t/(10))+c_{2}sin(t/(10)))