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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d y}{d x} = y + 7

\frac{d x}{d t} = 14 - 0.28 \cdot x

\frac{d x}{d t} = \frac{8 \cdot 1 0 ^{- 3}}{100} - 8 \frac{x}{9500}

\frac{d y}{d t} = \frac{3 t ^{2}}{8 y}

y^{^{'}} = (8 x - 5) (y + 2)

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