What is the solution for (22+x^2)y^'-2xy=0 ?

The solution for (22+x^2)y^'-2xy=0 is y=22e^{c_{1}}+e^{c_{1}}x^2

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

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

y = 22 e^{c_{1}} + e^{c_{1}} x^{2}

Solution steps

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

Solve separable ODE:

y = 22 e^{c_{1}} + e^{c_{1}} x^{2}

y = 22 e^{c_{1}} + e^{c_{1}} x^{2}

\int \frac{x ^{2}}{( x ^{3} + 1 )} d x

\frac{d r}{d θ} = π sin (π θ), r (1) = 3

(- \frac{4}{3}) \cdot e^{- 2 t} + (- \frac{2}{3}) \cdot e^{t}

cos (t + 1)

\int \frac{x ^{2} + x}{x ^{2} + x + 1} d x

What is the solution for (22+x^2)y^'-2xy=0 ?
The solution for (22+x^2)y^'-2xy=0 is y=22e^{c_{1}}+e^{c_{1}}x^2