What is the solution for y^{''}+4y=cos(2t),y(0)=1,y^'(0)=1 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y=cos(2t),y(0)=1,y^'(0)=1

y^{^{''}} + 4 y = cos (2 t), y (0) = 1, y^{^{'}} (0) = 1

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

Solution steps

y^{''} (t) + 4 y = cos (2 t)

Solve linear ODE:

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

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

\frac{d ^{2} x}{d t ^{2}} + 2 \frac{d x}{d t} + x = 3 cos (t)

y^{^{''}} - 18 y = tan (x)

8 y^{^{''}} + 0.25 y^{^{'}} + 512 y = 2 cos (2 t)

y^{^{''}} + y = 1, 6 cos (t) y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - y = 15 t, y (0) = - 15 t

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