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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(5 t + 2 y + 6) d t - d y = 0

\frac{d}{d x} (\frac{d y}{d x}) + 1 = 0

\int \frac{x ^{3} + 1}{x ^{3}} \frac{1}{x ^{4}} d x

f (y) = ln (y)

\int_{0}^{l n (π)} 2 x e^{x^{2}} sin (e^{x^{2}}) d x

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