What is the solution for y^{''}+16y=-32sin(4x) ?

The solution for y^{''}+16y=-32sin(4x) is y=c_{1}cos(4x)+c_{2}sin(4x)+4xcos(4x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+16y=-32sin(4x)

y^{^{''}} + 16 y = - 32 sin (4 x)

y = c_{1} cos (4 x) + c_{2} sin (4 x) + 4 x cos (4 x)

Solution steps

y^{''} (x) + 16 y = - 32 sin (4 x)

Solve linear ODE:

y = c_{1} cos (4 x) + c_{2} sin (4 x) + 4 x cos (4 x)

y = c_{1} cos (4 x) + c_{2} sin (4 x) + 4 x cos (4 x)

f (x) = 9 x - x^{2}, (1, 8)

\frac{s + 3}{s ( s + 2 )}

x \to 2 lim (\frac{2 x - 4}{3 - 2 x})

\frac{d}{d x} (ln tan (\frac{x}{2}) + C)

\frac{d ^{2}}{d x ^{2}} (3 sec (x))

What is the solution for y^{''}+16y=-32sin(4x) ?
The solution for y^{''}+16y=-32sin(4x) is y=c_{1}cos(4x)+c_{2}sin(4x)+4xcos(4x)