What is the solution for (5y-2x)y^'+2y=0 ?

The solution for (5y-2x)y^'+2y=0 is y=-(2x)/(5\W(-\frac{2x){c_{1}})}

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(5y-2x)y^'+2y=0

(5 y - 2 x) y^{^{'}} + 2 y = 0

y = - \frac{2 x}{5 W ( - \frac{2 x}{c _{1}} )}

Solution steps

(5 y - 2 x) y^{'} (x) + 2 y = 0

Solve using substitution:

y = - \frac{2 x}{5 W ( - \frac{2 x}{c _{1}} )}

y = - \frac{2 x}{5 W ( - \frac{2 x}{c _{1}} )}

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What is the solution for (5y-2x)y^'+2y=0 ?
The solution for (5y-2x)y^'+2y=0 is y=-(2x)/(5\W(-\frac{2x){c_{1}})}