What is the solution for y^{''}+3y^'-4y=2,y^'(0)=0,y(0)=0 ?

The solution for y^{''}+3y^'-4y=2,y^'(0)=0,y(0)=0 is y= 2/5 e^t+1/10 e^{-4t}-1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+3y^'-4y=2,y^'(0)=0,y(0)=0

y^{^{''}} + 3 y^{^{'}} - 4 y = 2, y^{^{'}} (0) = 0, y (0) = 0

y = \frac{2}{5} e^{t} + \frac{1}{10} e^{- 4 t} - \frac{1}{2}

Solution steps

y^{''} (t) + 3 y^{'} (t) - 4 y = 2

Solve linear ODE:

y = \frac{2}{5} e^{t} + \frac{1}{10} e^{- 4 t} - \frac{1}{2}

y = \frac{2}{5} e^{t} + \frac{1}{10} e^{- 4 t} - \frac{1}{2}

y^{^{''}} - 3 y^{^{'}} - 4 y = 3 e^{2} t + 2 sin (t)

y^{^{''}} + 3 y^{^{'}} - 4 y = x^{2} e^{x}

y^{^{''}} + 4 y^{^{'}} = 4 cos (x) + 3 sin (x) - 8

y^{^{''}} - y^{^{'}} - 6 y = e^{- x} + 12 x

5 y^{^{''}} - 15 y = 40 e^{x}

What is the solution for y^{''}+3y^'-4y=2,y^'(0)=0,y(0)=0 ?
The solution for y^{''}+3y^'-4y=2,y^'(0)=0,y(0)=0 is y= 2/5 e^t+1/10 e^{-4t}-1/2