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

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

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(x^2+1)(dy)/(dx)+3x(y-1)=0,y(0)=9

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

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

Solution steps

(x^{2} + 1) \frac{d y}{d 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

\frac{\partial}{\partial y} (2 x^{2} + 4 y^{2} + \frac{1}{2})

y^{^{'}} = 1 + y

\int_{0}^{1} \frac{1}{x - 1} d x

y^{^{'}} = \frac{( x y + 3 x - y - 3 )}{( x y - 2 x + 4 y - 8 )}

\frac{4}{( 3 s - 1 ) ^{5}}

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