What is the solution for (D^2+D-2)y=2x-40cos(2x) ?

The solution for (D^2+D-2)y=2x-40cos(2x) is y=c_{1}e^x+c_{2}e^{-2x}-x-1/2-2sin(2x)+6cos(2x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(D^2+D-2)y=2x-40cos(2x)

(D^{2} + D - 2) y = 2 x - 40 cos (2 x)

y = c_{1} e^{x} + c_{2} e^{- 2 x} - x - \frac{1}{2} - 2 sin (2 x) + 6 cos (2 x)

Solution steps

(D^{2} + D - 2) y = 2 x - 40 cos (2 x)

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 2 x} - x - \frac{1}{2} - 2 sin (2 x) + 6 cos (2 x)

y = c_{1} e^{x} + c_{2} e^{- 2 x} - x - \frac{1}{2} - 2 sin (2 x) + 6 cos (2 x)

y^{^{''''}} + 16 y^{^{''}} = sin (4 t) + t e^{5 t} + 4

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

y^{^{''}} + 9 y = t^{2} e^{3 t} + 6

y^{^{'''}} + 2 y^{^{''}} - 13 y^{^{'}} + 10 y = x e^{- x}

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

What is the solution for (D^2+D-2)y=2x-40cos(2x) ?
The solution for (D^2+D-2)y=2x-40cos(2x) is y=c_{1}e^x+c_{2}e^{-2x}-x-1/2-2sin(2x)+6cos(2x)