What is the solution for y^{''}+y^'-6y=0,y(0)=α,y^'(0)=5 ?

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

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)=α,y^'(0)=5

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = α, y^{^{'}} (0) = 5

y = e^{2 t} (α + \frac{5 - 2 α}{5}) - \frac{e ^{- 3 t} ( 5 - 2 α )}{5}

Solution steps

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

Solve linear ODE:

y = e^{2 t} (α + \frac{5 - 2 α}{5}) - \frac{e ^{- 3 t} ( 5 - 2 α )}{5}

y = e^{2 t} (α + \frac{5 - 2 α}{5}) - \frac{e ^{- 3 t} ( 5 - 2 α )}{5}

49 x^{^{''}} + 140 x^{^{'}} + 100 x = 0

35 y^{^{''}} + 23 y^{^{'}} - 6 y = 0

y^{^{''}} + 25 y = 10 y

y^{^{''}} - 2 y^{^{'}} + 2 y = 0, y (π) = 6, y (0) = 6

y^{^{''}} + 9 y = 0, y (0) = 0, y (1) = 0

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