What is the solution for xy^'+y=40sqrt(x) ?

The solution for xy^'+y=40sqrt(x) is y=(80x^{1/2})/3+(c_{1})/x

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xy^'+y=40sqrt(x)

x y^{^{'}} + y = 40 x

y = \frac{80 x ^{\frac{1}{2}}}{3} + \frac{c _{1}}{x}

Solution steps

x y^{'} (x) + y = 40 x

Solve linear ODE:

y = \frac{80 x ^{\frac{1}{2}}}{3} + \frac{c _{1}}{x}

y = \frac{80 x ^{\frac{1}{2}}}{3} + \frac{c _{1}}{x}

What is the solution for xy^'+y=40sqrt(x) ?
The solution for xy^'+y=40sqrt(x) is y=(80x^{1/2})/3+(c_{1})/x