What is the solution for y^'+1/x y= 7/(x^2)+3 ?

The solution for y^'+1/x y= 7/(x^2)+3 is y=(7ln(x))/x+(3x)/2+(c_{1})/x

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y^'+1/x y= 7/(x^2)+3

y^{^{'}} + \frac{1}{x} y = \frac{7}{x ^{2}} + 3

y = \frac{7 ln ( x )}{x} + \frac{3 x}{2} + \frac{c _{1}}{x}

Solution steps

y^{'} (x) + \frac{1}{x} y = \frac{7}{x ^{2}} + 3

Solve linear ODE:

y = \frac{7 ln ( x )}{x} + \frac{3 x}{2} + \frac{c _{1}}{x}

y = \frac{7 ln ( x )}{x} + \frac{3 x}{2} + \frac{c _{1}}{x}

x \to 1 lim (\frac{x - 3}{x ^{2} - 9})

\frac{d}{d x} (lo g_{4} (x^{2} - 9 x))

\int \frac{3}{- 1 - x} d x

x \to 0 lim (\frac{1}{2} + x^{2})

\int 9 x^{2} sin (π x) d x

What is the solution for y^'+1/x y= 7/(x^2)+3 ?
The solution for y^'+1/x y= 7/(x^2)+3 is y=(7ln(x))/x+(3x)/2+(c_{1})/x