What is the solution for y^{''}+16y=cos(4t),y(0)=3,y^'(0)=2 ?

The solution for y^{''}+16y=cos(4t),y(0)=3,y^'(0)=2 is y=3cos(4t)+1/2 sin(4t)+(tsin(4t))/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+16y=cos(4t),y(0)=3,y^'(0)=2

y^{^{''}} + 16 y = cos (4 t), y (0) = 3, y^{^{'}} (0) = 2

y = 3 cos (4 t) + \frac{1}{2} sin (4 t) + \frac{t sin ( 4 t )}{8}

Solution steps

y^{''} (t) + 16 y = cos (4 t)

Solve linear ODE:

y = 3 cos (4 t) + \frac{1}{2} sin (4 t) + \frac{t sin ( 4 t )}{8}

y = 3 cos (4 t) + \frac{1}{2} sin (4 t) + \frac{t sin ( 4 t )}{8}

y^{^{''}} - 5 y^{^{'}} - 6 y = 10

3 \frac{d ^{2} y}{d x ^{2}} + 20 \frac{d y}{d x} + 25 y = cos (2 x)

y^{^{'''}} + 3 y^{^{'}} - 2 x = 0

(D^{2} + 4 D + 5) y = e^{(- x)} + 15 x

y^{^{''}} + y = 2 cos (x), y^{^{'}} (π) = 0, y (0) = 0

What is the solution for y^{''}+16y=cos(4t),y(0)=3,y^'(0)=2 ?
The solution for y^{''}+16y=cos(4t),y(0)=3,y^'(0)=2 is y=3cos(4t)+1/2 sin(4t)+(tsin(4t))/8