What is the solution for y^{''}-y^'-20y=0,y(0)=α,y^'(0)=2 ?

The solution for y^{''}-y^'-20y=0,y(0)=α,y^'(0)=2 is y=e^{5t}(α+(2-5α)/9)-(e^{-4t}(2-5α))/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y^'-20y=0,y(0)=α,y^'(0)=2

y^{^{''}} - y^{^{'}} - 20 y = 0, y (0) = α, y^{^{'}} (0) = 2

y = e^{5 t} (α + \frac{2 - 5 α}{9}) - \frac{e ^{- 4 t} ( 2 - 5 α )}{9}

Solution steps

y^{''} (t) - y^{'} (t) - 20 y = 0

Solve linear ODE:

y = e^{5 t} (α + \frac{2 - 5 α}{9}) - \frac{e ^{- 4 t} ( 2 - 5 α )}{9}

y = e^{5 t} (α + \frac{2 - 5 α}{9}) - \frac{e ^{- 4 t} ( 2 - 5 α )}{9}

144 y^{^{''}} - 312 y^{^{'}} + 169 y = 0

y^{^{''}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} - 18 y^{^{'}} + 97 y = 0

\frac{d ^{3} y}{d x ^{3}} + 3 \frac{d ^{2} y}{d x ^{2}} = 0

y^{^{''}} + \frac{16}{3} y = 0

What is the solution for y^{''}-y^'-20y=0,y(0)=α,y^'(0)=2 ?
The solution for y^{''}-y^'-20y=0,y(0)=α,y^'(0)=2 is y=e^{5t}(α+(2-5α)/9)-(e^{-4t}(2-5α))/9