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

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

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

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

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

Solution steps

(x^{2} + 1) y^{'} (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{d}{d x} (x^{2} - 6 x + 8)

\frac{2 x}{1 + x ^{2}}

x \to 0 - lim (1)

n = 1 \sum \infty ln (n + 1)

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