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}}

\int \frac{x ^{2} + 25}{9 x ^{2}} d x

x \to 1 + lim (\frac{x + 5}{x + 2})

\frac{\partial}{\partial x} (ln (4 y))

y^{^{''}} - 5 y^{^{'}} + 4 y = - sin (5 t)

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

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)