What is the solution for x^2y^'-xy=x^{-7}y^{1/2} ?

The solution for x^2y^'-xy=x^{-7}y^{1/2} is y= 1/(289x^{16)}-(2c_{1})/(17x^{15/2)}+c_{1}^2x

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x^2y^'-xy=x^{-7}y^{1/2}

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

y = \frac{1}{289 x ^{16}} - \frac{2 c _{1}}{17 x ^{\frac{15}{2}}} + c_{1}^{2} x

Solution steps

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

Solve Bernoulli ODE:

y = \frac{1}{289 x ^{16}} - \frac{2 c _{1}}{17 x ^{\frac{15}{2}}} + c_{1}^{2} x

y = \frac{1}{289 x ^{16}} - \frac{2 c _{1}}{17 x ^{\frac{15}{2}}} + c_{1}^{2} x

x \to 2 lim (arctan (\frac{( 2 x ^{2} - 8 )}{3 x ^{2} - 6 x}))

y = - x^{2} - 2 x, y = - 2 x - 4, x = - 3, x = 0

y^{^{'''}} + y^{^{'}} = tan (x)

\frac{d}{d x} (6 sec (x) - 5 x)

x \to - 1 lim ((x + 3 x^{3})^{4})

What is the solution for x^2y^'-xy=x^{-7}y^{1/2} ?
The solution for x^2y^'-xy=x^{-7}y^{1/2} is y= 1/(289x^{16)}-(2c_{1})/(17x^{15/2)}+c_{1}^2x