What is the solution for x^2(y^')^1-3xy^'+16y=0 ?

The solution for x^2(y^')^1-3xy^'+16y=0 is y=e^{-16(-1/3 ln(x)+1/3 ln(x-3))+c_{1}}

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x^2(y^')^1-3xy^'+16y=0

x^{2} (y^{^{'}})^{1} - 3 x y^{^{'}} + 16 y = 0

y = e^{- 16 (- \frac{1}{3} l n (x) + \frac{1}{3} l n (x - 3)) + c_{1}}

Solution steps

x^{2} (y^{'} (x))^{1} - 3 x y^{'} (x) + 16 y = 0

Solve separable ODE:

y = e^{- 16 (- \frac{1}{3} l n (x) + \frac{1}{3} l n (x - 3)) + c_{1}}

y = e^{- 16 (- \frac{1}{3} l n (x) + \frac{1}{3} l n (x - 3)) + c_{1}}

What is the solution for x^2(y^')^1-3xy^'+16y=0 ?
The solution for x^2(y^')^1-3xy^'+16y=0 is y=e^{-16(-1/3 ln(x)+1/3 ln(x-3))+c_{1}}