What is the solution for y^{''}+4y^'+4y=2x+6,y(0)=1,y^'(0)= 1/2 ?

The solution for y^{''}+4y^'+4y=2x+6,y(0)=1,y^'(0)= 1/2 is y= x/2+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+4y^'+4y=2x+6,y(0)=1,y^'(0)= 1/2

y^{^{''}} + 4 y^{^{'}} + 4 y = 2 x + 6, y (0) = 1, y^{^{'}} (0) = \frac{1}{2}

y = \frac{x}{2} + 1

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = 2 x + 6

Solve linear ODE:

y = \frac{x}{2} + 1

y = \frac{x}{2} + 1

What is the solution for y^{''}+4y^'+4y=2x+6,y(0)=1,y^'(0)= 1/2 ?
The solution for y^{''}+4y^'+4y=2x+6,y(0)=1,y^'(0)= 1/2 is y= x/2+1