What is the solution for y^{''}+y=3sin(4x) ?

The solution for y^{''}+y=3sin(4x) is y=c_{1}cos(x)+c_{2}sin(x)-1/5 sin(4x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y=3sin(4x)

y^{^{''}} + y = 3 sin (4 x)

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{5} sin (4 x)

Solution steps

y^{''} (x) + y = 3 sin (4 x)

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{5} sin (4 x)

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{5} sin (4 x)

f (x) = \frac{( x + 1 ) ( 2 x - 3 )}{2 x + 3}

y^{^{''}} + 2 y^{^{'}} - 3 y = - 16 x e^{x}

y = 3 - x, y = 2 x^{2}, [0, 3]

\frac{d}{d x} (4 x + 2 y)

\frac{\partial}{\partial x} (- 6 x^{3} + y^{2} + 5 x y)

What is the solution for y^{''}+y=3sin(4x) ?
The solution for y^{''}+y=3sin(4x) is y=c_{1}cos(x)+c_{2}sin(x)-1/5 sin(4x)