What is the solution for (x^2+1)y^'+3xy=6x ?

The solution for (x^2+1)y^'+3xy=6x is y=(c_{1})/((x^2+1)^{3/2)}+2

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(x^2+1)y^'+3xy=6x

(x^{2} + 1) y^{^{'}} + 3 x y = 6 x

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

Solution steps

(x^{2} + 1) y^{'} (x) + 3 x y = 6 x

Solve separable ODE:

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

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

\int sin^{3} (3 x) cos^{6} (3 x) d x

\int x ln (5 x) d x

\int 2 x + y^{2} d x

x \to 8 lim (x^{4} \cdot e^{- x^{2}})

y = 10 - (x + 2)^{2}, (- 5.2, 1.2)

What is the solution for (x^2+1)y^'+3xy=6x ?
The solution for (x^2+1)y^'+3xy=6x is y=(c_{1})/((x^2+1)^{3/2)}+2