What is the solution for x^3y^'+3x^2y= 1/x ?

The solution for x^3y^'+3x^2y= 1/x is y=(ln(x))/(x^3)+(c_{1})/(x^3)

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x^3y^'+3x^2y= 1/x

x^{3} y^{^{'}} + 3 x^{2} y = \frac{1}{x}

y = \frac{ln ( x )}{x ^{3}} + \frac{c _{1}}{x ^{3}}

Solution steps

x^{3} y^{'} (x) + 3 x^{2} y = \frac{1}{x}

Solve linear ODE:

y = \frac{ln ( x )}{x ^{3}} + \frac{c _{1}}{x ^{3}}

y = \frac{ln ( x )}{x ^{3}} + \frac{c _{1}}{x ^{3}}

\frac{d}{d x} (5 x^{2} sin (x) + 4 x)

f (x) = \frac{x ^{4} + c _{1}}{4 x ^{2}}

\int 55 - 6 x - x^{2} d x

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