What is the solution for (x^2+1)y^'+5x(y-1)=0 ?

The solution for (x^2+1)y^'+5x(y-1)=0 is y=(c_{1})/((x^2+1)^{5/2)}+1

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(x^2+1)y^'+5x(y-1)=0

(x^{2} + 1) y^{^{'}} + 5 x (y - 1) = 0

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

Solution steps

(x^{2} + 1) y^{'} (x) + 5 x (y - 1) = 0

Solve separable ODE:

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

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

\int_{- 1}^{3} x + 1 d x

\frac{\partial}{\partial y} (2 x^{3} y + 3)

\int_{- 3}^{- 5} - 4^{- 3} d x

\frac{d}{d x} (e^{- 16 x})

n = 1 \sum \infty \frac{6}{π ^{n}}

What is the solution for (x^2+1)y^'+5x(y-1)=0 ?
The solution for (x^2+1)y^'+5x(y-1)=0 is y=(c_{1})/((x^2+1)^{5/2)}+1