What is the solution for (dy)/(dx)= 1/2 (kx^{-1}-x) ?

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

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(dy)/(dx)= 1/2 (kx^{-1}-x)

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

y = \frac{1}{2} (k ln (x) - \frac{x ^{2}}{2}) + c_{1}

Solution steps

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

Substitute

\frac{d y}{d x}

with

y^{'} (x)

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

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

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

\int \frac{1}{2} (k x^{- 1} - x) d x = \frac{1}{2} (k ln (x) - \frac{x ^{2}}{2}) + c_{1}

y = \frac{1}{2} (k ln (x) - \frac{x ^{2}}{2}) + c_{1}

x \to - 4 lim (- \frac{2 x ^{2} - 17}{3})

\int 5 y^{2} d y

\frac{\partial}{\partial x} (2 cos (x) cos (y))

h \to 0 lim (\frac{13 h}{h})

x \to 2 lim (2 x^{2} - 7 x - 15)

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