What is the solution for y^{''}-y^'-2y=8x+6 ?

The solution for y^{''}-y^'-2y=8x+6 is y=c_{1}e^{2x}+c_{2}e^{-x}-4x-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y^'-2y=8x+6

y^{^{''}} - y^{^{'}} - 2 y = 8 x + 6

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

Solution steps

y^{''} (x) - y^{'} (x) - 2 y = 8 x + 6

Solve linear ODE:

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

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

y^{^{''}} - y = \frac{2 e ^{x}}{e ^{x} - 1}

y^{^{''}} + 7 y + 6 y = e^{6 x}

9 y^{^{''}} - 6 y^{^{'}} + y = 8

y^{^{'''}} - 9 y^{^{''}} + 27 y^{^{'}} - 27 y = 5 e^{3 x}

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

What is the solution for y^{''}-y^'-2y=8x+6 ?
The solution for y^{''}-y^'-2y=8x+6 is y=c_{1}e^{2x}+c_{2}e^{-x}-4x-1