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

The solution for (x^2+1)y^'+6x(y-1)=0,y(0)=6 is y= 5/((x^2+1)^3)+1

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

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

y = \frac{5}{( x ^{2} + 1 ) ^{3}} + 1

Solution steps

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

Solve separable ODE:

y = \frac{5}{( x ^{2} + 1 ) ^{3}} + 1

y = \frac{5}{( x ^{2} + 1 ) ^{3}} + 1

ln (x + 1)

\frac{d ^{3}}{d x ^{3}} (sin (7 x))

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

f (x) = 3 - x - x^{2}, at x = 1

\frac{d}{d x} (x^{3} + 4 x^{2} - 6 x + 1)

What is the solution for (x^2+1)y^'+6x(y-1)=0,y(0)=6 ?
The solution for (x^2+1)y^'+6x(y-1)=0,y(0)=6 is y= 5/((x^2+1)^3)+1