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

The solution for y^{''}+y^'=cos(x),y(0)=0,y^'(0)=2 is y= 3/2+1/2-3/2 e^{-x}+(sin(x)-cos(x))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = \frac{3}{2} + \frac{1}{2} - \frac{3}{2} e^{- x} + \frac{sin ( x ) - cos ( x )}{2}

Solution steps

y^{''} (x) + y^{'} (x) = cos (x)

Solve linear ODE:

y = \frac{3}{2} + \frac{1}{2} - \frac{3}{2} e^{- x} + \frac{sin ( x ) - cos ( x )}{2}

y = \frac{3}{2} + \frac{1}{2} - \frac{3}{2} e^{- x} + \frac{sin ( x ) - cos ( x )}{2}

y^{^{''}} - 9 y = cosh (3 x)

y^{^{''''}} - 16 y = - 12 t^{3}

y^{^{''''}} - 16 y = 75 sin (t)

y^{^{''}} - 100 y = sinh (10 x)

14 y^{^{''}} + y^{^{'}} + y = x^{2} - 5 x

What is the solution for y^{''}+y^'=cos(x),y(0)=0,y^'(0)=2 ?
The solution for y^{''}+y^'=cos(x),y(0)=0,y^'(0)=2 is y= 3/2+1/2-3/2 e^{-x}+(sin(x)-cos(x))/2