What is the solution for y^{''}+4y=3cos(2t),y(0)=1,y^'(0)=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y=3cos(2t),y(0)=1,y^'(0)=0

y^{^{''}} + 4 y = 3 cos (2 t), y (0) = 1, y^{^{'}} (0) = 0

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

Solution steps

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

Solve linear ODE:

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

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

0.125 y^{^{''}} + 106 y + 1000 = 0

\frac{d ^{2} y}{d x ^{2}} - y = e^{x} sin (x)

y^{^{''}} + y^{^{'}} = 1, y (0) = 0, y^{^{'}} (0) = 1

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

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = x^{2} + x

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