What is the solution for y^'+(2y)/x =-4/x ,y(1)=3 ?

The solution for y^'+(2y)/x =-4/x ,y(1)=3 is y= 5/(x^2)-2

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y^'+(2y)/x =-4/x ,y(1)=3

y^{^{'}} + \frac{2 y}{x} = - \frac{4}{x}, y (1) = 3

y = \frac{5}{x ^{2}} - 2

Solution steps

y^{'} (x) + \frac{2 y}{x} = - \frac{4}{x}

Solve separable ODE:

y = \frac{5}{x ^{2}} - 2

y = \frac{5}{x ^{2}} - 2

(\frac{1}{e ^{x}})^{^{'}}

\frac{d}{d x} (x + x + 8)

(5 x + y e^{\frac{y}{x}}) d x - x e^{\frac{y}{x}} d y = 0

y^{^{'}} = \frac{y}{x} + (\frac{2 x ^{3}}{y}) cos (x^{2})

x \to 0 lim ((2 x - 1)^{3})

What is the solution for y^'+(2y)/x =-4/x ,y(1)=3 ?
The solution for y^'+(2y)/x =-4/x ,y(1)=3 is y= 5/(x^2)-2