What is the solution for y^{''}+16y=4sin(2x) ?

The solution for y^{''}+16y=4sin(2x) is y=c_{1}cos(4x)+c_{2}sin(4x)+1/3 sin(2x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+16y=4sin(2x)

y^{^{''}} + 16 y = 4 sin (2 x)

y = c_{1} cos (4 x) + c_{2} sin (4 x) + \frac{1}{3} sin (2 x)

Solution steps

y^{''} (x) + 16 y = 4 sin (2 x)

Solve linear ODE:

y = c_{1} cos (4 x) + c_{2} sin (4 x) + \frac{1}{3} sin (2 x)

y = c_{1} cos (4 x) + c_{2} sin (4 x) + \frac{1}{3} sin (2 x)

\frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} - 15 y = 6 + 4 t

5 y^{^{''}} + y = sin (2 x) + x

(4 D^{2} + 4 D + 1) y = x e^{- \frac{x}{2}} sin (x)

y^{^{''}} + 2 y^{^{'}} + 4 y = 111 e^{2 x} cos (3 x)

y^{^{''''}} - 4 y^{^{''}} = x^{2} + e^{x}

What is the solution for y^{''}+16y=4sin(2x) ?
The solution for y^{''}+16y=4sin(2x) is y=c_{1}cos(4x)+c_{2}sin(4x)+1/3 sin(2x)