What is the solution for y^'+y-3cos(x)=0 ?

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

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y^'+y-3cos(x)=0

y^{^{'}} + y - 3 cos (x) = 0

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

Solution steps

y^{'} (x) + y - 3 cos (x) = 0

Solve linear ODE:

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

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

\frac{d y}{d x}, x^{5} + y^{5} = 35 x y

\frac{\partial}{\partial y} (e^{- 1 x} cos (6 y))

\frac{e ^{x}}{- 8 x ^{7}}

\frac{d}{d x} (\frac{x ^{5}}{e ^{x}})

\frac{d}{d x} (- \frac{24}{x ^{5}})

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