What is the solution for y^{''}-4y=0,y(0)=0,y^'(0)=3 ?

The solution for y^{''}-4y=0,y(0)=0,y^'(0)=3 is y= 3/4 e^{2t}-3/4 e^{-2t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y=0,y(0)=0,y^'(0)=3

y^{^{''}} - 4 y = 0, y (0) = 0, y^{^{'}} (0) = 3

y = \frac{3}{4} e^{2 t} - \frac{3}{4} e^{- 2 t}

Solution steps

y^{''} (t) - 4 y = 0

Solve linear ODE:

y = \frac{3}{4} e^{2 t} - \frac{3}{4} e^{- 2 t}

y = \frac{3}{4} e^{2 t} - \frac{3}{4} e^{- 2 t}

y^{^{'''}} - 7 y^{^{''}} + y^{^{'}} - 7 y = 0

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

y^{^{''}} + 9 y^{^{'}} + 12 y = 0

3 y^{^{''}} - 16 y^{^{'}} + 13 y = 0

y^{^{''}} + n^{2} y = 0, y (0) = c, y^{^{'}} (0) = v

What is the solution for y^{''}-4y=0,y(0)=0,y^'(0)=3 ?
The solution for y^{''}-4y=0,y(0)=0,y^'(0)=3 is y= 3/4 e^{2t}-3/4 e^{-2t}