What is the solution for x^2y^'+2xy=13cos^2(x) ?

The solution for x^2y^'+2xy=13cos^2(x) is y=(13(2x+sin(2x)))/(4x^2)+(c_{1})/(x^2)

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x^2y^'+2xy=13cos^2(x)

x^{2} y^{^{'}} + 2 x y = 13 cos^{2} (x)

y = \frac{13 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

Solution steps

x^{2} y^{'} (x) + 2 x y = 13 cos^{2} (x)

Solve linear ODE:

y = \frac{13 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

y = \frac{13 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

y^{^{'}} + 3 x^{2} y = 6 x^{2}

\int - cot (2 x) d x

\int x (a - x)^{2} d x

\frac{d}{d x} (\frac{2}{x ^{2} + 1})

\int_{0}^{2} 2 π (5 - x) (8 - x^{3}) d x

What is the solution for x^2y^'+2xy=13cos^2(x) ?
The solution for x^2y^'+2xy=13cos^2(x) is y=(13(2x+sin(2x)))/(4x^2)+(c_{1})/(x^2)