What is the solution for (x^2+1)y^'+8x(y-1)=0,y(0)=3 ?

The solution for (x^2+1)y^'+8x(y-1)=0,y(0)=3 is y= 2/((x^2+1)^4)+1

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(x^2+1)y^'+8x(y-1)=0,y(0)=3

(x^{2} + 1) y^{^{'}} + 8 x (y - 1) = 0, y (0) = 3

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

Solution steps

(x^{2} + 1) y^{'} (x) + 8 x (y - 1) = 0

Solve separable ODE:

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

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

\frac{\partial}{\partial x} (4 x^{2} - 4 x y + x yz)

\int \frac{tan ( x )}{1 + cos ( x )} d x

6 - x^{2}, x^{2} - 2, [0, 3]

\int \frac{ln ( y )}{ln ( y )} d y

e^{t^{2}} y^{^{'}} + (t + t y^{2}) = 0, y (0) = 1

What is the solution for (x^2+1)y^'+8x(y-1)=0,y(0)=3 ?
The solution for (x^2+1)y^'+8x(y-1)=0,y(0)=3 is y= 2/((x^2+1)^4)+1