What is the solution for y^{''}+y^'-6y=0,y(1)=2e^2-e^{-3} ?

The solution for y^{''}+y^'-6y=0,y(1)=2e^2-e^{-3} is y=e^{2t-5}(2e^5-1-c_{2})+c_{2}e^{-3t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y^'-6y=0,y(1)=2e^2-e^{-3}

y^{^{''}} + y^{^{'}} - 6 y = 0, y (1) = 2 e^{2} - e^{- 3}

y = e^{2 t - 5} (2 e^{5} - 1 - c_{2}) + c_{2} e^{- 3 t}

Solution steps

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

Solve linear ODE:

y = e^{2 t - 5} (2 e^{5} - 1 - c_{2}) + c_{2} e^{- 3 t}

y = e^{2 t - 5} (2 e^{5} - 1 - c_{2}) + c_{2} e^{- 3 t}

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

8 y^{^{''}} - 9 y^{^{'}} + y = 0

y^{^{'''}} + 4 y^{^{''}} + y^{^{'}} - 26 y = 0

y^{^{''}} - 2 y^{^{'}} + 5 y = 0

y^{^{''}} - 2 y^{^{'}} + 2 y = 0, y^{^{'}} (π) = 1, y (0) = 1

What is the solution for y^{''}+y^'-6y=0,y(1)=2e^2-e^{-3} ?
The solution for y^{''}+y^'-6y=0,y(1)=2e^2-e^{-3} is y=e^{2t-5}(2e^5-1-c_{2})+c_{2}e^{-3t}