What is the solution for y^{''}+k^2y=cos(kx) ?

The solution for y^{''}+k^2y=cos(kx) is y=c_{1}cos(kx)+c_{2}sin(kx)+(xsin(kx))/(2k)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+k^2y=cos(kx)

y^{^{''}} + k^{2} y = cos (k x)

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

Solution steps

y^{''} (x) + k^{2} y = cos (k x)

Solve linear ODE:

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

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

\int_{0}^{2} \frac{12}{x ^{2} + 4} d x

\frac{\partial}{\partial y} (y - x^{2})

\int 3^{3}

\int (\frac{1}{x} - 1)^{2} d x

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

What is the solution for y^{''}+k^2y=cos(kx) ?
The solution for y^{''}+k^2y=cos(kx) is y=c_{1}cos(kx)+c_{2}sin(kx)+(xsin(kx))/(2k)