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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 25 y = e^{5 x}, y (0) = 1, y^{^{'}} (0) = - 4

x^{^{''}} + 25 x = t (sin (5 t) + cos (5 t))

4 y^{^{''}} - y = e^{(\frac{x}{2})} + 5

4 y^{^{''}} - y = e^{(\frac{x}{2})} + 7

- y^{^{''}} + y = e^{- 1 \cdot x}, y (0) = 0, y (1) = 0

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