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

The solution for y^'+4y=cos(x) is y=(sin(x))/(17)+(4cos(x))/(17)+(c_{1})/(e^{4x)}

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y^'+4y=cos(x)

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

y = \frac{sin ( x )}{17} + \frac{4 cos ( x )}{17} + \frac{c _{1}}{e ^{4 x}}

Solution steps

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

Solve linear ODE:

y = \frac{sin ( x )}{17} + \frac{4 cos ( x )}{17} + \frac{c _{1}}{e ^{4 x}}

y = \frac{sin ( x )}{17} + \frac{4 cos ( x )}{17} + \frac{c _{1}}{e ^{4 x}}

a re a 6 x - 18, [2, 6]

\int - 9 csc^{2} (x) d x

d er i v a t i v e g (x) = 9 - x^{2}

t an g e n t f (x) = - 4 - x^{2}

\int_{- \infty}^{0} \frac{1}{3 - 9 x} d x

What is the solution for y^'+4y=cos(x) ?
The solution for y^'+4y=cos(x) is y=(sin(x))/(17)+(4cos(x))/(17)+(c_{1})/(e^{4x)}