What is the solution for (x^2+1)(dy)/(dx)+4x(y-1)=0,y(0)=6 ?

The solution for (x^2+1)(dy)/(dx)+4x(y-1)=0,y(0)=6 is y= 5/((x^2+1)^2)+1

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(x^2+1)(dy)/(dx)+4x(y-1)=0,y(0)=6

(x^{2} + 1) \frac{d y}{d x} + 4 x (y - 1) = 0, y (0) = 6

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

Solution steps

(x^{2} + 1) \frac{d y}{d x} + 4 x (y - 1) = 0

Solve separable ODE:

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

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

y = 2 x + 4

7 x - 8 y x^{2} + 1 (\frac{d y}{d x}) = 0

\int 9 x + \frac{1}{9 x} d x

\frac{d ^{2} y}{d x ^{2}} - 4 y = 0, y (0) = 2

\frac{d}{d x} (ln (e^{4 x}))

What is the solution for (x^2+1)(dy)/(dx)+4x(y-1)=0,y(0)=6 ?
The solution for (x^2+1)(dy)/(dx)+4x(y-1)=0,y(0)=6 is y= 5/((x^2+1)^2)+1