What is the solution for x^2y^'+4xy=(sin(3x))/(x^2) ?

The solution for x^2y^'+4xy=(sin(3x))/(x^2) is y=-(cos(3x))/(3x^4)+(c_{1})/(x^4)

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x^2y^'+4xy=(sin(3x))/(x^2)

x^{2} y^{^{'}} + 4 x y = \frac{sin ( 3 x )}{x ^{2}}

y = - \frac{cos ( 3 x )}{3 x ^{4}} + \frac{c _{1}}{x ^{4}}

Solution steps

x^{2} y^{'} (x) + 4 x y = \frac{sin ( 3 x )}{x ^{2}}

Solve linear ODE:

y = - \frac{cos ( 3 x )}{3 x ^{4}} + \frac{c _{1}}{x ^{4}}

y = - \frac{cos ( 3 x )}{3 x ^{4}} + \frac{c _{1}}{x ^{4}}

\int \frac{2 x - 1}{x - 1} d x

\int \frac{1}{t ^{2} 9 - t ^{2}} d t

\frac{d ^{2}}{d y ^{2}} (cos (x y^{2}))

\int 5 x ln (3 x) d x

f (x) = arctan (\frac{1}{x ^{2}})

What is the solution for x^2y^'+4xy=(sin(3x))/(x^2) ?
The solution for x^2y^'+4xy=(sin(3x))/(x^2) is y=-(cos(3x))/(3x^4)+(c_{1})/(x^4)