What is the solution for x^2y^'+2xy=9cos^2(x) ?

The solution for x^2y^'+2xy=9cos^2(x) is y=(9(2x+sin(2x)))/(4x^2)+(c_{1})/(x^2)

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x^2y^'+2xy=9cos^2(x)

x^{2} y^{^{'}} + 2 x y = 9 cos^{2} (x)

y = \frac{9 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

Solution steps

x^{2} y^{'} (x) + 2 x y = 9 cos^{2} (x)

Solve linear ODE:

y = \frac{9 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

y = \frac{9 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

\frac{\partial}{\partial t} (x^{2} sin (3 t))

\frac{d}{d x} (\frac{( x ^{2} - 1 )}{2 x - 3})

x \to 1 + lim (∣ x ∣)

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

y = \frac{3}{2}, y = \frac{1}{1 - x ^{2}}

What is the solution for x^2y^'+2xy=9cos^2(x) ?
The solution for x^2y^'+2xy=9cos^2(x) is y=(9(2x+sin(2x)))/(4x^2)+(c_{1})/(x^2)