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

The solution for y^'(x^2+1)+3x(y-1)=0 is y=e^{-(3x^2)/2+c_{1}}+1

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

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

y = e^{- \frac{3 x ^{2}}{2} + c_{1}} + 1

Solution steps

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

Solve separable ODE:

y = e^{- \frac{3 x ^{2}}{2} + c_{1}} + 1

y = e^{- \frac{3 x ^{2}}{2} + c_{1}} + 1

What is the solution for y^'(x^2+1)+3x(y-1)=0 ?
The solution for y^'(x^2+1)+3x(y-1)=0 is y=e^{-(3x^2)/2+c_{1}}+1