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

The solution for y^{''}+2y=e^x+2 is y=c_{1}cos(sqrt(2)x)+c_{2}sin(sqrt(2)x)+1/3 e^x+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{3} e^{x} + 1

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{3} e^{x} + 1

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{3} e^{x} + 1

y^{^{''}} + y^{^{'}} - 2 y = 3 e^{2 x}

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 6 e^{2 x} - 2

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

y^{^{''}} + 2 y = 8 x + 5

y^{^{''}} + 9 y^{^{'}} = cos (3 t)

What is the solution for y^{''}+2y=e^x+2 ?
The solution for y^{''}+2y=e^x+2 is y=c_{1}cos(sqrt(2)x)+c_{2}sin(sqrt(2)x)+1/3 e^x+1