What is the solution for x^2y^'+2xy=11cos^2(x) ?

The solution for x^2y^'+2xy=11cos^2(x) is y=(11(2x+sin(2x)))/(4x^2)+(c_{1})/(x^2)

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x^2y^'+2xy=11cos^2(x)

x^{2} y^{^{'}} + 2 x y = 11 cos^{2} (x)

y = \frac{11 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

Solution steps

x^{2} y^{'} (x) + 2 x y = 11 cos^{2} (x)

Solve linear ODE:

y = \frac{11 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

y = \frac{11 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

y = lo g_{5} (7 x + 1)

\frac{d}{d x} (\frac{4 x ^{3}}{3})

\frac{\partial}{\partial y} (z cos (x z))

\frac{d}{d x} ((x - 2)^{4} - 1)

\int x I n (1 + x)

What is the solution for x^2y^'+2xy=11cos^2(x) ?
The solution for x^2y^'+2xy=11cos^2(x) is y=(11(2x+sin(2x)))/(4x^2)+(c_{1})/(x^2)