What is the solution for y^{''}+3y^'+2y=4x+3 ?

The solution for y^{''}+3y^'+2y=4x+3 is y=c_{1}e^{-x}+c_{2}e^{-2x}+2x-3/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + 2 x - \frac{3}{2}

Solution steps

y^{''} (x) + 3 y^{'} (x) + 2 y = 4 x + 3

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + 2 x - \frac{3}{2}

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + 2 x - \frac{3}{2}

(d^{2} + d - 2) y = lo g_{6} (x)

(d^{2} + 5 d + 6) \cdot y = e^{x}

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

y^{^{''}} + 25 y = cos^{5} (x)

y^{^{'''}} + 3 y^{^{''}} + 3 y^{^{'}} + y = 30 e^{- x}

What is the solution for y^{''}+3y^'+2y=4x+3 ?
The solution for y^{''}+3y^'+2y=4x+3 is y=c_{1}e^{-x}+c_{2}e^{-2x}+2x-3/2