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

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

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(x^2+1)y^'+3x(y-1)=0,y(0)=9

(x^{2} + 1) y^{^{'}} + 3 x (y - 1) = 0, y (0) = 9

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

Solution steps

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

Solve separable ODE:

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

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

f (x) = x^{2} + 2 x - 3

x \to \frac{π}{2} lim (f (x))

\int 5^{3 - 2 x} d x

y^{^{''}} + 2 y^{^{'}} + y = x

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

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