What is the solution for y^'+y+2sin(x)=0 ?

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

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y^'+y+2sin(x)=0

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

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

Solution steps

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

Solve linear ODE:

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

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

e^{2 x} + 2 x

\frac{d}{d x} (\frac{1}{3 8 x ^{2}})

\int_{1}^{e} ln (8 x) d x

\frac{d}{d x} (\frac{1}{2 x - 3})

f (x) = \frac{3 x + 2}{x + 1}

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