What is the solution for (x^2+4)y^'+xy=9x ?

The solution for (x^2+4)y^'+xy=9x is y=(c_{1})/(sqrt(x^2+4))+9

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(x^2+4)y^'+xy=9x

(x^{2} + 4) y^{^{'}} + x y = 9 x

y = \frac{c _{1}}{x ^{2} + 4} + 9

Solution steps

(x^{2} + 4) y^{'} (x) + x y = 9 x

Solve separable ODE:

y = \frac{c _{1}}{x ^{2} + 4} + 9

y = \frac{c _{1}}{x ^{2} + 4} + 9

y (1 + x^{3}) y^{^{'}} + x^{2} (1 + y^{2}) = 0

\frac{d y}{d x} = \frac{1 - y}{e ^{0.1 x}}

\frac{x ( d y )}{( d x ) - 4 y} = 2 x^{(5)} e^{(x)}

\frac{d x}{x} = - v^{2} d v

2 y y^{^{'}} = x (x^{2} - 16)^{(\frac{- 1}{2})}

What is the solution for (x^2+4)y^'+xy=9x ?
The solution for (x^2+4)y^'+xy=9x is y=(c_{1})/(sqrt(x^2+4))+9