What is the solution for y^{''}+y^'=2sin(x) ?

The solution for y^{''}+y^'=2sin(x) is y=c_{1}+c_{2}e^{-x}-cos(x)-sin(x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y^'=2sin(x)

y^{^{''}} + y^{^{'}} = 2 sin (x)

y = c_{1} + c_{2} e^{- x} - cos (x) - sin (x)

Solution steps

y^{''} (x) + y^{'} (x) = 2 sin (x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} - cos (x) - sin (x)

y = c_{1} + c_{2} e^{- x} - cos (x) - sin (x)

y^{^{''}} + 6 y^{^{'}} + 9 y = 11 e^{- 3 x}

y^{^{''}} - 5 y^{^{'}} + 4 y = 8 e^{- 2 x}

y \cdot u^{2} - y^{^{''}} = 1

y^{^{''}} + 4 y^{^{'}} + 4 y = x^{2} e - 2 x + 1

y^{^{''}} - y^{^{'}} = 1 - k x

What is the solution for y^{''}+y^'=2sin(x) ?
The solution for y^{''}+y^'=2sin(x) is y=c_{1}+c_{2}e^{-x}-cos(x)-sin(x)