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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

\int_{0}^{1} x x^{2} + 1 d x

\int \frac{1}{7 x ^{7}} d x

\frac{d}{d x} (sin (4 x) + 2 x^{\frac{3}{2}})

sin (x), x = \frac{π}{4}

\frac{d}{d x} ((ln (x)) d x)

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