What is the solution for (dy)/(dx)+(3/x)y=-x+4 ?

The solution for (dy)/(dx)+(3/x)y=-x+4 is y=-(x^2)/5+x+(c_{1})/(x^3)

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(dy)/(dx)+(3/x)y=-x+4

\frac{d y}{d x} + (\frac{3}{x}) y = - x + 4

y = - \frac{x ^{2}}{5} + x + \frac{c _{1}}{x ^{3}}

Solution steps

\frac{d y}{d x} + (\frac{3}{x}) y = - x + 4

Solve linear ODE:

y = - \frac{x ^{2}}{5} + x + \frac{c _{1}}{x ^{3}}

y = - \frac{x ^{2}}{5} + x + \frac{c _{1}}{x ^{3}}

\frac{\partial}{\partial x} (ln (\frac{1}{2 + 3 e ^{x}}))

x \to 1 lim (\frac{3 x - 2}{x - 1})

\frac{1}{3} tan^{3} (x) - tan (x) + x

x \to - 8 lim (\frac{x + 2}{x + 8})

x \to 2 lim (2 x^{2} + 7 x - 1)

What is the solution for (dy)/(dx)+(3/x)y=-x+4 ?
The solution for (dy)/(dx)+(3/x)y=-x+4 is y=-(x^2)/5+x+(c_{1})/(x^3)