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

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

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xy^'+2y=(sin(x))/x

x y^{^{'}} + 2 y = \frac{sin ( x )}{x}

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

Solution steps

x y^{'} (x) + 2 y = \frac{sin ( x )}{x}

Solve linear ODE:

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

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

y^{^{'}} = 7 x e^{y}

k \to 0 lim (\frac{x ^{2} k + 3 x k ^{2} + k ^{3}}{2 x k + 5 k ^{2}})

\frac{d}{d x} (\frac{1 + x}{1 + e ^{x}})

f (x) = sin (5 x)

x \to - 3 lim (- x^{2} + 6 x - 2)

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