What is the solution for (x^2+1)y^'+6x(y-1)=0,y(0)=9 ?

The solution for (x^2+1)y^'+6x(y-1)=0,y(0)=9 is y= 8/((x^2+1)^3)+1

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(x^2+1)y^'+6x(y-1)=0,y(0)=9

(x^{2} + 1) y^{^{'}} + 6 x (y - 1) = 0, y (0) = 9

y = \frac{8}{( x ^{2} + 1 ) ^{3}} + 1

Solution steps

(x^{2} + 1) y^{'} (x) + 6 x (y - 1) = 0

Solve separable ODE:

y = \frac{8}{( x ^{2} + 1 ) ^{3}} + 1

y = \frac{8}{( x ^{2} + 1 ) ^{3}} + 1

(x - 1) ln (x), 1

\frac{\partial}{\partial x} (sin (x))

x \to 0 lim (\frac{( 3 e ^{x} - 3 )}{t})

\int x^{2} + \frac{3}{4} d x

\int y sin (y^{2}) d y

What is the solution for (x^2+1)y^'+6x(y-1)=0,y(0)=9 ?
The solution for (x^2+1)y^'+6x(y-1)=0,y(0)=9 is y= 8/((x^2+1)^3)+1