What is the solution for y^{''}-y^'-6y=-e^x+12 ?

The solution for y^{''}-y^'-6y=-e^x+12 is y=c_{1}e^{3x}+c_{2}e^{-2x}+1/6 e^x-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y^'-6y=-e^x+12

y^{^{''}} - y^{^{'}} - 6 y = - e^{x} + 12

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} + \frac{1}{6} e^{x} - 2

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = - e^{x} + 12

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} + \frac{1}{6} e^{x} - 2

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} + \frac{1}{6} e^{x} - 2

y^{^{''}} + 4 y = sec (22 x)

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

\frac{d ^{2} y}{d x ^{2}} + y = (sec (x))^{3}

y^{^{''}} - 2 y^{^{'}} - 3 y = 2 e^{5 t}

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

What is the solution for y^{''}-y^'-6y=-e^x+12 ?
The solution for y^{''}-y^'-6y=-e^x+12 is y=c_{1}e^{3x}+c_{2}e^{-2x}+1/6 e^x-2