What is the solution for y^{''}-4y=0,y(0)=0,y^'(0)=6 ?

The solution for y^{''}-4y=0,y(0)=0,y^'(0)=6 is y= 3/2 e^{2t}-3/2 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)=6

y^{^{''}} - 4 y = 0, y (0) = 0, y^{^{'}} (0) = 6

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - \frac{4 y ^{^{'}}}{3} + \frac{97 y}{36} = 0

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = \frac{31}{2}

4 \frac{d y}{d x} - 3 \frac{d ^{2} y}{d x ^{2}} = 0

y^{^{'''}} + 5 y^{^{''}} - 11 y^{^{'}} - 39 y = 0

4 y^{^{''}} + 12 y^{^{'}} + 9 y = 0, y (0) = 3, y^{^{'}} (0) = 5

What is the solution for y^{''}-4y=0,y(0)=0,y^'(0)=6 ?
The solution for y^{''}-4y=0,y(0)=0,y^'(0)=6 is y= 3/2 e^{2t}-3/2 e^{-2t}