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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

y^{''} (x) - y^{'} (x) - 2 y = x

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + y = 2 cos (2 x) + 3 x + 2 + 3 e^{x}

y^{^{''}} + y = t^{2} sin (t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{'''}} - y^{^{'}} = t

x^{^{''}} + x = sin (2 t), x (0) = x, x (0) = 0, x^{^{'}} (0) = 0

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

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