What is the solution for y^{''}+3y^'+2y=0,y(0)= 1/4 ?

The solution for y^{''}+3y^'+2y=0,y(0)= 1/4 is y=(1/4-c_{2})e^{-t}+c_{2}e^{-2t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+3y^'+2y=0,y(0)= 1/4

y^{^{''}} + 3 y^{^{'}} + 2 y = 0, y (0) = \frac{1}{4}

y = (\frac{1}{4} - c_{2}) e^{- t} + c_{2} e^{- 2 t}

Solution steps

y^{''} (t) + 3 y^{'} (t) + 2 y = 0

Solve linear ODE:

y = (\frac{1}{4} - c_{2}) e^{- t} + c_{2} e^{- 2 t}

y = (\frac{1}{4} - c_{2}) e^{- t} + c_{2} e^{- 2 t}

y^{^{''}} + (3 - a) y^{^{'}} + 2 (a - 1) y = 0

\frac{d ^{2} y}{d x ^{2}} - 12 \frac{d y}{d x} - 64 y = 0

\frac{10}{32} y^{^{''}} + \frac{1}{6} y = 0, y (0) = \frac{1}{6}, y^{^{'}} (0) = - 1

10000 y^{^{''}} - 729 y = 0, y (0) = 9

y^{^{''}} + 16 y = 0, y (π^{8}) = - 4, y^{^{'}} (π^{8}) = - 8

What is the solution for y^{''}+3y^'+2y=0,y(0)= 1/4 ?
The solution for y^{''}+3y^'+2y=0,y(0)= 1/4 is y=(1/4-c_{2})e^{-t}+c_{2}e^{-2t}