What is the solution for x^2y^'+3xy=((sin(2x)))/x ?

The solution for x^2y^'+3xy=((sin(2x)))/x is y=-(cos(2x))/(2x^3)+(c_{1})/(x^3)

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x^2y^'+3xy=((sin(2x)))/x

x^{2} y^{^{'}} + 3 x y = \frac{( sin ( 2 x ) )}{x}

y = - \frac{cos ( 2 x )}{2 x ^{3}} + \frac{c _{1}}{x ^{3}}

Solution steps

x^{2} y^{'} (x) + 3 x y = \frac{( sin ( 2 x ) )}{x}

Solve linear ODE:

y = - \frac{cos ( 2 x )}{2 x ^{3}} + \frac{c _{1}}{x ^{3}}

y = - \frac{cos ( 2 x )}{2 x ^{3}} + \frac{c _{1}}{x ^{3}}

x \to 8 lim (\frac{x - 9}{x - 8})

f (x) = ((25 - x^{2})^{\frac{1}{2}}), \frac{4}{3} x, 5

\frac{\partial}{\partial x} (a e^{2 x})

\int_{0}^{2} \frac{( x + 2 )}{x - 4} d x

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

What is the solution for x^2y^'+3xy=((sin(2x)))/x ?
The solution for x^2y^'+3xy=((sin(2x)))/x is y=-(cos(2x))/(2x^3)+(c_{1})/(x^3)