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

The solution for xy^'+2y=2cos(x) is y=(2(xsin(x)+cos(x)))/(x^2)+(c_{1})/(x^2)

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xy^'+2y=2cos(x)

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{x ^{2} - 9 x + 1}{x ^{2} + 5}

5 sin^{2} (t)

\frac{d}{d x} (4 x sin (x^{2}))

\frac{d}{d x} (\frac{2 - x}{1 - y})

\frac{\partial}{\partial x} (e^{- \frac{x}{2}})

What is the solution for xy^'+2y=2cos(x) ?
The solution for xy^'+2y=2cos(x) is y=(2(xsin(x)+cos(x)))/(x^2)+(c_{1})/(x^2)