What is the solution for y^'-((2x)/(1+x^2))y=0,y(0)=4 ?

The solution for y^'-((2x)/(1+x^2))y=0,y(0)=4 is y=4(x^2+1)

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y^'-((2x)/(1+x^2))y=0,y(0)=4

y^{^{'}} - (\frac{2 x}{1 + x ^{2}}) y = 0, y (0) = 4

y = 4 (x^{2} + 1)

Solution steps

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

Solve separable ODE:

y = 4 (x^{2} + 1)

y = 4 (x^{2} + 1)

d er i v a t i v e (t^{2} - t) (t^{2} + t)

\frac{\partial}{\partial y} (e^{3 x y})

\frac{d y}{d x} + 5 y = 0

\frac{\partial}{\partial x} (5 + e^{- x y} cos (y))

x \to 7 lim (\frac{x ^{2} - 5 x - 14}{x - 7})

What is the solution for y^'-((2x)/(1+x^2))y=0,y(0)=4 ?
The solution for y^'-((2x)/(1+x^2))y=0,y(0)=4 is y=4(x^2+1)