What is the solution for x^2y^'+4xy=(sin(2x))/(x^2) ?

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

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x^2y^'+4xy=(sin(2x))/(x^2)

x^{2} y^{^{'}} + 4 x y = \frac{sin ( 2 x )}{x ^{2}}

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

Solution steps

x^{2} y^{'} (x) + 4 x y = \frac{sin ( 2 x )}{x ^{2}}

Solve linear ODE:

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

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

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