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

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

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

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

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

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

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

r^{2} - 16 x^{2}

\frac{\partial}{\partial x} (y - x^{2})

\frac{\partial}{\partial x} (r cos (x))

x \to 7 lim (3 x - 4)

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