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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (arcsin (x) d x)

f (x) = (x^{3} - 25 x)^{8}, (- 5, 0)

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

\int \frac{- 3 x}{( x - 1 ) ^{2} + 4} d x

y^{^{'''}} - y^{^{''}} + y^{^{'}} + 3 y = 0

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