What is the solution for x^2y^'+2xy=5cos^2(x) ?

The solution for x^2y^'+2xy=5cos^2(x) is y=(5(2x+sin(2x)))/(4x^2)+(c_{1})/(x^2)

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x^2y^'+2xy=5cos^2(x)

x^{2} y^{^{'}} + 2 x y = 5 cos^{2} (x)

y = \frac{5 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

Solution steps

x^{2} y^{'} (x) + 2 x y = 5 cos^{2} (x)

Solve linear ODE:

y = \frac{5 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

y = \frac{5 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

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

y = \frac{6 x ^{2} + 6 x + 4}{x}

f (x) = \frac{3 x}{x + 1}, at x = 2

\frac{4 x}{x ^{2} + 1}

What is the solution for x^2y^'+2xy=5cos^2(x) ?
The solution for x^2y^'+2xy=5cos^2(x) is y=(5(2x+sin(2x)))/(4x^2)+(c_{1})/(x^2)