What is the solution for y^{''}+y-2=5e^{-2x} ?

The solution for y^{''}+y-2=5e^{-2x} is y=c_{1}cos(x)+c_{2}sin(x)+e^{-2x}+2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

y^{''} (x) + y - 2 = 5 e^{- 2 x}

Solve linear ODE:

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

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

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

y^{^{''}} - 6 y^{^{'}} + 9 y = 4 e^{3} t ln (t)

y^{^{''}} + y = cos (2 t) cos (t)

y^{^{''}} - 7 y^{^{'}} + 12 y = - e^{x}

y^{^{''}} - 2 y^{^{'}} + y = e^{t} + t, y^{^{'}} (0) = 0, y (0) = 1

What is the solution for y^{''}+y-2=5e^{-2x} ?
The solution for y^{''}+y-2=5e^{-2x} is y=c_{1}cos(x)+c_{2}sin(x)+e^{-2x}+2