What is the solution for y^'-1/2 y=(5x-5)y^{-1},y(0)=4 ?

The solution for y^'-1/2 y=(5x-5)y^{-1},y(0)=4 is y=sqrt(-10x+16e^x)

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y^'-1/2 y=(5x-5)y^{-1},y(0)=4

y^{^{'}} - \frac{1}{2} y = (5 x - 5) y^{- 1}, y (0) = 4

y = - 10 x + 16 e^{x}

Solution steps

y^{'} (x) - \frac{1}{2} y = (5 x - 5) y^{- 1}

Solve Bernoulli ODE:

y = - 10 x + 16 e^{x}

y = - 10 x + 16 e^{x}

\int \frac{x ^{5}}{1 - x ^{4}} d x

\int (5 x + 5) (x - 4) (x - 4) d x

\int 9 t^{2} + \frac{1}{t ^{3}} d t

y^{^{''}} + y^{^{'}} - 2 y = 6 e^{x}

4 y^{^{''}} + 8 y^{^{'}} + 3 y = 0

What is the solution for y^'-1/2 y=(5x-5)y^{-1},y(0)=4 ?
The solution for y^'-1/2 y=(5x-5)y^{-1},y(0)=4 is y=sqrt(-10x+16e^x)