What is the solution for x^2y^'+2xy=21cos^2(x) ?

The solution for x^2y^'+2xy=21cos^2(x) is y=(21(2x+sin(2x)))/(4x^2)+(c_{1})/(x^2)

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x^2y^'+2xy=21cos^2(x)

x^{2} y^{^{'}} + 2 x y = 21 cos^{2} (x)

y = \frac{21 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

Solution steps

x^{2} y^{'} (x) + 2 x y = 21 cos^{2} (x)

Solve linear ODE:

y = \frac{21 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

y = \frac{21 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

\frac{x}{x - 1}

\frac{d}{d x} (x^{9 x})

\int_{0}^{1} (x^{2} - 8) e^{- x} d x

(x^{2} - y^{2}) \frac{d y}{d x} = 2 x y

x^{2} y y^{^{'}} = (y^{2} - 1)^{\frac{3}{2}}

What is the solution for x^2y^'+2xy=21cos^2(x) ?
The solution for x^2y^'+2xy=21cos^2(x) is y=(21(2x+sin(2x)))/(4x^2)+(c_{1})/(x^2)