What is the solution for y^'+5y=cos(x) ?

The solution for y^'+5y=cos(x) is y=(sin(x))/(26)+(5cos(x))/(26)+(c_{1})/(e^{5x)}

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y^'+5y=cos(x)

y^{^{'}} + 5 y = cos (x)

y = \frac{sin ( x )}{26} + \frac{5 cos ( x )}{26} + \frac{c _{1}}{e ^{5 x}}

Solution steps

y^{'} (x) + 5 y = cos (x)

Solve linear ODE:

y = \frac{sin ( x )}{26} + \frac{5 cos ( x )}{26} + \frac{c _{1}}{e ^{5 x}}

y = \frac{sin ( x )}{26} + \frac{5 cos ( x )}{26} + \frac{c _{1}}{e ^{5 x}}

\int 3 x sec^{2} (4 x^{2}) d x

\int e^{x} + x^{2} + x^{3} d x

\int_{0}^{1} t e^{- s t} d t

\frac{d}{d x} (60 x^{4} + 180 x^{3} - 240 x^{2})

f (x) = (- 2 x^{2} - 4)^{3}

What is the solution for y^'+5y=cos(x) ?
The solution for y^'+5y=cos(x) is y=(sin(x))/(26)+(5cos(x))/(26)+(c_{1})/(e^{5x)}