What is the solution for derivative of f(x^{-1})x^2-8x-9x=-21 ?

The solution for derivative of f(x^{-1})x^2-8x-9x=-21 is f(x)=-21+(17x)/2+(c_{1})/x

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derivative of f(x^{-1})x^2-8x-9x=-21

\frac{d}{d x} (f (x)^{- 1}) x^{2} - 8 x - 9 x = - 21

f (x) = - 21 + \frac{17 x}{2} + \frac{c _{1}}{x}

Solution steps

\frac{d}{d x} ((f (x) x^{- 1}) x^{2}) - 8 x - 9 x = - 21

Isolate

(f (x) x^{- 1} x^{2})^{'} : (f (x) x^{- 1} x^{2})^{'} = - 21 + 17 x

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

then

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

f (x) x^{- 1} x^{2} = \int - 21 + 17 x d x

\int - 21 + 17 x d x = - 21 x + \frac{17 x ^{2}}{2} + c_{1}

f (x) x^{- 1} x^{2} = - 21 x + \frac{17 x ^{2}}{2} + c_{1}

Isolate

f (x) : f (x) = - 21 + \frac{17 x}{2} + \frac{c _{1}}{x}

f (x) = - 21 + \frac{17 x}{2} + \frac{c _{1}}{x}

y^{^{'}} = \frac{y - 1}{x}

\frac{d y}{d x} = - 2 (y - 40)

3 + 2 q + (4 + 2 p) \frac{d q}{d p} = 0

sin (3 x) d x + 2 y cos (33 x) d y = 0

cos (x) y^{^{'}} = \frac{y}{ln ( y )}, y (0) = 1

What is the solution for derivative of f(x^{-1})x^2-8x-9x=-21 ?
The solution for derivative of f(x^{-1})x^2-8x-9x=-21 is f(x)=-21+(17x)/2+(c_{1})/x