What is the solution for y^{''}-y-12y=0 ?

The solution for y^{''}-y-12y=0 is y=c_{1}e^{sqrt(13)t}+c_{2}e^{-sqrt(13)t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y-12y=0

y^{^{''}} - y - 12 y = 0

y = c_{1} e^{13 t} + c_{2} e^{- 13 t}

Solution steps

y^{''} (t) - y - 12 y = 0

Solve linear ODE:

y = c_{1} e^{13 t} + c_{2} e^{- 13 t}

y = c_{1} e^{13 t} + c_{2} e^{- 13 t}

f^{^{''}} (x) + f^{^{'}} (x) = 0

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

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

\frac{d ^{2} y}{d x ^{2}} + 16 \frac{d y}{d x} + 64 y = 0

y^{^{''}} + 14 y^{^{'}} + 64 y = 0, y (0) = 5, y^{^{'}} (0) = 0

What is the solution for y^{''}-y-12y=0 ?
The solution for y^{''}-y-12y=0 is y=c_{1}e^{sqrt(13)t}+c_{2}e^{-sqrt(13)t}