What is the solution for y^{''}+y=-3cos(x)sin(x) ?

The solution for y^{''}+y=-3cos(x)sin(x) is y=c_{1}cos(x)+c_{2}sin(x)+cos(x)sin(x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y=-3cos(x)sin(x)

y^{^{''}} + y = - 3 cos (x) sin (x)

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

Solution steps

y^{''} (x) + y = - 3 cos (x) sin (x)

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} - 3 y = - 9 cos (2 x) - 33 sin (2 x)

y^{^{''}} - 8 y^{^{'}} + 16 y = 12 x + 6

(d^{2} + 4) y = cos (2 x) + cos (4 x)

y^{^{'''}} - y^{^{''}} - 25 y^{^{'}} + 25 y = 9 - e^{x} + e^{5 x}

y^{^{''}} + 4 y^{^{'}} + 13 y = t e^{- t}, y (0) = 0, y^{^{'}} (0) = 4

What is the solution for y^{''}+y=-3cos(x)sin(x) ?
The solution for y^{''}+y=-3cos(x)sin(x) is y=c_{1}cos(x)+c_{2}sin(x)+cos(x)sin(x)