What is the solution for xy^'+y=25sqrt(x) ?

The solution for xy^'+y=25sqrt(x) is y=(50x^{1/2})/3+(c_{1})/x

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xy^'+y=25sqrt(x)

x y^{^{'}} + y = 25 x

y = \frac{50 x ^{\frac{1}{2}}}{3} + \frac{c _{1}}{x}

Solution steps

x y^{'} (x) + y = 25 x

Solve linear ODE:

y = \frac{50 x ^{\frac{1}{2}}}{3} + \frac{c _{1}}{x}

y = \frac{50 x ^{\frac{1}{2}}}{3} + \frac{c _{1}}{x}

d y - (y - 4)^{2} d x = 0

\frac{d}{d x} (cos^{2} (x s i) n^{2} x)

\frac{\partial}{\partial x} (x arcsin (x y))

f (x) = \frac{6}{x}

\frac{\partial}{\partial x} ((x - y)^{5})

What is the solution for xy^'+y=25sqrt(x) ?
The solution for xy^'+y=25sqrt(x) is y=(50x^{1/2})/3+(c_{1})/x