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}}

\int_{4}^{6} \frac{1}{x} d x

x \to - 1 lim (x^{3} - 4 x + 1)

n = 10000 \sum \infty \frac{1}{n}

\int \frac{x - arctan ( 2 x )}{1 + 4 x ^{2}} d x

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)