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

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

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)=0

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

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

Solution steps

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

Solve linear ODE:

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

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

36 y^{^{''}} + 156 y^{^{'}} + 169 y = 0, y (0) = 9, y^{^{'}} (0) = 4

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

y^{^{''}} - 2 y + y = 0, y (0) = x e^{x}

\frac{d ^{2} y}{d x ^{2}} + 10 \frac{d y}{d x} + 61 y = 0

77 y^{^{''''}} + 219 y^{^{''}} + 100 y = 0

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