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

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

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y^'+y-2cos(x)=0

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

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

Solution steps

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

Solve linear ODE:

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

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

t^{10} e^{- 7 t}

\int sin (\frac{nπ x}{3}) d x

n = 0 \sum \infty \frac{1}{n + 3}

\int \frac{1}{2 x ^{5}} d x

x \to 0 lim (4 - x^{2})

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