What is the solution for y^'+y-4cos(x)=0 ?

The solution for y^'+y-4cos(x)=0 is y=2(sin(x)+cos(x))+(c_{1})/(e^x)

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y^'+y-4cos(x)=0

y^{^{'}} + y - 4 cos (x) = 0

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

Solution steps

y^{'} (x) + y - 4 cos (x) = 0

Solve linear ODE:

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

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

\frac{s}{( s + 2 ) ( s + 3 )}

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

x \to 0 lim (\frac{1 - cos ( x )}{5 x ^{2}})

y^{^{'}} - y = 10 t e^{2 t}

\int \frac{1}{y ^{2} + 1} d y

What is the solution for y^'+y-4cos(x)=0 ?
The solution for y^'+y-4cos(x)=0 is y=2(sin(x)+cos(x))+(c_{1})/(e^x)