What is the solution for y^{''}-y^'=4cos(2x),y(0)=1,y^'(0)=0 ?

The solution for y^{''}-y^'=4cos(2x),y(0)=1,y^'(0)=0 is y=(-1+9/5)e^x+1-2/5 sin(2x)-4/5 cos(2x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-y^'=4cos(2x),y(0)=1,y^'(0)=0

y^{^{''}} - y^{^{'}} = 4 cos (2 x), y (0) = 1, y^{^{'}} (0) = 0

y = (- 1 + \frac{9}{5}) e^{x} + 1 - \frac{2}{5} sin (2 x) - \frac{4}{5} cos (2 x)

Solution steps

y^{''} (x) - y^{'} (x) = 4 cos (2 x)

Solve linear ODE:

y = (- 1 + \frac{9}{5}) e^{x} + 1 - \frac{2}{5} sin (2 x) - \frac{4}{5} cos (2 x)

y = (- 1 + \frac{9}{5}) e^{x} + 1 - \frac{2}{5} sin (2 x) - \frac{4}{5} cos (2 x)

What is the solution for y^{''}-y^'=4cos(2x),y(0)=1,y^'(0)=0 ?
The solution for y^{''}-y^'=4cos(2x),y(0)=1,y^'(0)=0 is y=(-1+9/5)e^x+1-2/5 sin(2x)-4/5 cos(2x)