What is the solution for (x^2+5)y^'+xy-6x=0 ?

The solution for (x^2+5)y^'+xy-6x=0 is y=(c_{1})/(sqrt(x^2+5))+6

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(x^2+5)y^'+xy-6x=0

(x^{2} + 5) y^{^{'}} + x y - 6 x = 0

y = \frac{c _{1}}{x ^{2} + 5} + 6

Solution steps

(x^{2} + 5) y^{'} (x) + x y - 6 x = 0

Solve separable ODE:

y = \frac{c _{1}}{x ^{2} + 5} + 6

y = \frac{c _{1}}{x ^{2} + 5} + 6

\frac{d y}{d x} y = e^{6 x} sin (x)

y^{^{'}} + cy = 0

\frac{d y}{d x} = y^{2} (sin (x) + x)

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

(1 - x^{2}) \frac{d y}{d x} - 2 x \frac{d y}{d x} + 3 y = 0

What is the solution for (x^2+5)y^'+xy-6x=0 ?
The solution for (x^2+5)y^'+xy-6x=0 is y=(c_{1})/(sqrt(x^2+5))+6