What is the solution for (d^2y)/(dx^2)+4y=0,y=-cos(2x) ?

The solution for (d^2y)/(dx^2)+4y=0,y=-cos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(d^2y)/(dx^2)+4y=0,y=-cos(2x)

\frac{d ^{2} y}{d x ^{2}} + 4 y = 0, y = - cos (2 x)

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

Solution steps

\frac{d ^{2} y}{d x ^{2}} + 4 y = 0

Solve linear ODE:

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

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

y^{^{'''}} + 2 y^{^{''}} - 11 y^{^{'}} - 12 y = 0

y^{^{''}} - 5 y^{^{'}} - 6 y = 0, y (0) = x, y^{^{'}} (0) = 4

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = - 2

y^{^{''}} + 20 y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = - 24

y^{^{'''}} - 3 y^{^{''}} - 54 y^{^{'}} = 0

What is the solution for (d^2y)/(dx^2)+4y=0,y=-cos(2x) ?
The solution for (d^2y)/(dx^2)+4y=0,y=-cos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)