What is the solution for y^{''}+y^'-6y=0,y(0)=7,y^'(0)=4 ?

The solution for y^{''}+y^'-6y=0,y(0)=7,y^'(0)=4 is y=5e^{2t}+2e^{-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(0)=7,y^'(0)=4

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 7, y^{^{'}} (0) = 4

y = 5 e^{2 t} + 2 e^{- 3 t}

Solution steps

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

Solve linear ODE:

y = 5 e^{2 t} + 2 e^{- 3 t}

y = 5 e^{2 t} + 2 e^{- 3 t}

y^{^{''}} + 0 y^{^{'}} + 36 y = 0, y (0) = 1

y^{^{''}} + 20 y^{^{'}} + 24 y = 0, y (0) = 0, y^{^{'}} (0) = 0.6

4 y^{^{''}} + 577 y = 0, y (0) = 1, y^{^{'}} (0) = 3

2 y^{^{''}} + 4 y^{^{'}} + k y = 0, y (0) = 2, y^{^{'}} (0) = 0

0.1 \cdot y^{^{''}} + y^{^{'}} + y = 0

What is the solution for y^{''}+y^'-6y=0,y(0)=7,y^'(0)=4 ?
The solution for y^{''}+y^'-6y=0,y(0)=7,y^'(0)=4 is y=5e^{2t}+2e^{-3t}