What is the solution for y^{''}+4y=6cos(x) ?

The solution for y^{''}+4y=6cos(x) is y=c_{1}cos(2x)+c_{2}sin(2x)+2cos(x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y=6cos(x)

y^{^{''}} + 4 y = 6 cos (x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + 2 cos (x)

Solution steps

y^{''} (x) + 4 y = 6 cos (x)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + 2 cos (x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + 2 cos (x)

y^{^{''}} + 9 y^{^{'}} = e^{t}, y (0) = 0, y^{^{'}} (0) = 0

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

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

y^{^{''}} + 4 y^{^{'}} + 5 y = 5

4 y^{^{''}} + 4 y^{^{'}} + y = \frac{5 e ^{- \frac{t}{2}}}{t ^{2} + 1}

What is the solution for y^{''}+4y=6cos(x) ?
The solution for y^{''}+4y=6cos(x) is y=c_{1}cos(2x)+c_{2}sin(2x)+2cos(x)