The value of the y is [tex]y=\frac{x^8}{16}+\frac{c_1}{x^8}[/tex]
We have given that,
[tex]\frac{dy}{dx}=\frac{x^8-8y}{x}[/tex]
We have solved linear ordinary differential equation
The linear differential equation is of the form dy/dx + Py = Q, where P and Q are numeric constants or functions in x. It consists of a y and a derivative of y.
[tex]\frac{dy}{dx}=\frac{x^8-8y}{x}\\y'=\frac{x^8-8y}{x}\\\\y'+\frac{8y}{x} =x^8[/tex]
[tex]y=uv\\y'=uv'+vu'\\y=\frac{x^8}{16}+\frac{c_1}{x^8}[/tex]
The value of the y is [tex]y=\frac{x^8}{16}+\frac{c_1}{x^8}[/tex]
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