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

The solution for (x^2+1)(dy)/(dx)+3x(y-1)=0,y(0)=5 is y= 4/((x^2+1)^{3/2)}+1

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(x^2+1)(dy)/(dx)+3x(y-1)=0,y(0)=5

(x^{2} + 1) \frac{d y}{d x} + 3 x (y - 1) = 0, y (0) = 5

y = \frac{4}{( 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{4}{( x ^{2} + 1 ) ^{\frac{3}{2}}} + 1

y = \frac{4}{( x ^{2} + 1 ) ^{\frac{3}{2}}} + 1

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

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

(t^{2} - 16 t + 55) \cdot \frac{d y}{d t} = y

\frac{d y}{d x} = e^{- x - y - 2}

\frac{d y}{d x} = \frac{y}{( x + 1 ) x}

What is the solution for (x^2+1)(dy)/(dx)+3x(y-1)=0,y(0)=5 ?
The solution for (x^2+1)(dy)/(dx)+3x(y-1)=0,y(0)=5 is y= 4/((x^2+1)^{3/2)}+1