What is the solution for y^{''}-4y=0,y(0)=2,y^'(0)=1 ?

The solution for y^{''}-4y=0,y(0)=2,y^'(0)=1 is y=(2-3/4)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)=2,y^'(0)=1

y^{^{''}} - 4 y = 0, y (0) = 2, y^{^{'}} (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 13 y^{^{'}} + 40 y = 0

y^{^{'''}} + 6 y^{^{''}} + 13 y^{^{'}} = 0

0.2 y^{^{''}} + 5 y^{^{'}} + \frac{1}{0.002} y = 0

y^{^{''}} + 3 y^{^{'}} + a y = 0, y (0) = 1, y^{^{'}} (0) = - 8

y^{^{'''}} + 6 y^{^{''}} - 15 y^{^{'}} - 100 y = 0

What is the solution for y^{''}-4y=0,y(0)=2,y^'(0)=1 ?
The solution for y^{''}-4y=0,y(0)=2,y^'(0)=1 is y=(2-3/4)e^{2t}+3/4 e^{-2t}