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

The solution for (x^2+1)y^'=-3x(y-1) is y=(c_{1})/((x^2+1)^{3/2)}+1

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

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

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

Solution steps

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

Solve separable ODE:

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

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

- \frac{1}{2 x ^{4}}

\int_{0}^{18} \frac{1}{30} d x

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

\frac{d}{d t} (sin (9 t))

What is the solution for (x^2+1)y^'=-3x(y-1) ?
The solution for (x^2+1)y^'=-3x(y-1) is y=(c_{1})/((x^2+1)^{3/2)}+1