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

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

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(x^2+1)y^'+8x(y-1)=0,y(0)=4

(x^{2} + 1) y^{^{'}} + 8 x (y - 1) = 0, y (0) = 4

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

Solution steps

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

Solve separable ODE:

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

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

\int x e^{\frac{- x}{2}} d x

\int_{0}^{1} 8 ln (x) d x

\int (1 + e^{x}) d x

n = 1 \sum \infty \frac{8}{( n + 1 ) !}

\int (10 - x) d x

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