What is the solution for (dy)/(dx)x^2+yx=1,y(6)=9 ?

The solution for (dy)/(dx)x^2+yx=1,y(6)=9 is y=(ln(x))/x+(54-ln(6))/x

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(dy)/(dx)x^2+yx=1,y(6)=9

\frac{d y}{d x} x^{2} + y x = 1, y (6) = 9

y = \frac{ln ( x )}{x} + \frac{54 - ln ( 6 )}{x}

Solution steps

\frac{d y}{d x} x^{2} + y x = 1

Solve linear ODE:

y = \frac{ln ( x )}{x} + \frac{54 - ln ( 6 )}{x}

y = \frac{ln ( x )}{x} + \frac{54 - ln ( 6 )}{x}

\int e^{e} d e

x (\frac{d y}{d x}) = 1 - y^{2}

x \to \infty lim ((\frac{x + 1}{x + 3})^{\frac{x}{5}})

\frac{d y}{d x} + 9 y = e^{x} y^{- 4}

y^{^{''}} + 81 y = sin (9 t)

What is the solution for (dy)/(dx)x^2+yx=1,y(6)=9 ?
The solution for (dy)/(dx)x^2+yx=1,y(6)=9 is y=(ln(x))/x+(54-ln(6))/x