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

The solution for (x^2+1)y^'+4x(y-1)=0 is y=(c_{1})/(x^4+2x^2+1)+1

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

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

y = \frac{c _{1}}{x ^{4} + 2 x ^{2} + 1} + 1

Solution steps

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

Solve separable ODE:

y = \frac{c _{1}}{x ^{4} + 2 x ^{2} + 1} + 1

y = \frac{c _{1}}{x ^{4} + 2 x ^{2} + 1} + 1

49 - x^{2}

f (x) = x^{3} - 3 x^{2} + 4 x + 5, x = 3

(1 + x)^{- \frac{7}{3}}, 4

y^{^{''}} - 64 y = 0

x^{2} - 18, 17 x

What is the solution for (x^2+1)y^'+4x(y-1)=0 ?
The solution for (x^2+1)y^'+4x(y-1)=0 is y=(c_{1})/(x^4+2x^2+1)+1