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=(-c_{2}+1/2)e^t+c_{2}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 = (- c_{2} + \frac{1}{2}) e^{t} + c_{2} e^{- 4 t} - \frac{1}{2}

Solution steps

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

Solve linear ODE:

y = (- c_{2} + \frac{1}{2}) e^{t} + c_{2} e^{- 4 t} - \frac{1}{2}

y = (- c_{2} + \frac{1}{2}) e^{t} + c_{2} e^{- 4 t} - \frac{1}{2}

4 y^{^{''}} + 4 y^{^{'}} + 1 = 0

D (x) = a - 2

(D^{3} + 1) y = 3 + e^{- x} + 5 e^{2 x}

x^{^{''}} + 49 x = 39 cos (6 t)

4 y^{^{''}} - 8 y^{^{'}} + 4 y = e^{x} (3 x + 2) + 2 cos (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=(-c_{2}+1/2)e^t+c_{2}e^{-4t}-1/2