What is the solution for y^{''}-2y^'-3y=0,y(0)=2,y^'(0)=2 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = 0

Solve linear ODE:

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

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

(y + 4) y^{^{'}} = 2 x y + 4 y

\frac{1}{X - 1}, 12

\frac{d}{d x} (2 x^{3} - 2 x)

\int x sin (s) d x

y = (7) (6^{x - 8})

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