What is the solution for (x^2+1)(dy)/(dx)+3x(y-1)=0,y(0)=6 ?

The solution for (x^2+1)(dy)/(dx)+3x(y-1)=0,y(0)=6 is y= 5/((x^2+1)^{3/2)}+1

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(x^2+1)(dy)/(dx)+3x(y-1)=0,y(0)=6

(x^{2} + 1) \frac{d y}{d x} + 3 x (y - 1) = 0, y (0) = 6

y = \frac{5}{( 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{5}{( x ^{2} + 1 ) ^{\frac{3}{2}}} + 1

y = \frac{5}{( x ^{2} + 1 ) ^{\frac{3}{2}}} + 1

\frac{d}{d x} (\frac{d ^{2}}{d x ^{2}} (\frac{1}{x}) - ln (x))

x \to 0 lim ((\frac{4}{x} - 1) x)

5 x + x^{2}

\frac{d}{d x} (x (x - 4)^{\frac{1}{3}})

\frac{d y}{d t} = 3 t^{2}

What is the solution for (x^2+1)(dy)/(dx)+3x(y-1)=0,y(0)=6 ?
The solution for (x^2+1)(dy)/(dx)+3x(y-1)=0,y(0)=6 is y= 5/((x^2+1)^{3/2)}+1