What is the solution for (x^2+1)y^'+3x(y-1)=0,y(0)=4 ?

The solution for (x^2+1)y^'+3x(y-1)=0,y(0)=4 is y= 3/((x^2+1)^{3/2)}+1

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(x^2+1)y^'+3x(y-1)=0,y(0)=4

(x^{2} + 1) y^{^{'}} + 3 x (y - 1) = 0, y (0) = 4

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

Solution steps

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

Solve separable ODE:

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

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

\int \frac{x ^{3} + 1}{x ^{3} + x ^{2}} d x

y = e^{- x} arctan (5 e^{x})

x^{2} ln (x^{2} + 1)

\frac{d y}{d x}, y^{3} = x^{8}

\int \frac{( x + 6 ) ^{4}}{2 x} d x

What is the solution for (x^2+1)y^'+3x(y-1)=0,y(0)=4 ?
The solution for (x^2+1)y^'+3x(y-1)=0,y(0)=4 is y= 3/((x^2+1)^{3/2)}+1