What is the solution for (D^2-4)y=2cos(2x) ?

The solution for (D^2-4)y=2cos(2x) is y=c_{1}e^{2x}+c_{2}e^{-2x}-1/4 cos(2x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(D^2-4)y=2cos(2x)

(D^{2} - 4) y = 2 cos (2 x)

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

Solution steps

(D^{2} - 4) y = 2 cos (2 x)

Solve linear ODE:

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

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

y^{^{''}} + 9 y = cos (3 t), y (\frac{π}{6}) = 1, y (0) = 1

1.23 \cdot 2 0^{2} π 0.0 2^{2} - 0.017 y^{^{'}} = 0.01 y^{^{''}}

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

D (r) = \frac{27 ( r ) + 13}{26}

y^{^{''}} + 16 y = 4 (u \frac{7}{4} π)

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