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(1+x^2)(dy)/(dx)+2x y=1/(1+x^2); y=0 if...

`(1+x^2)(dy)/(dx)+2x y=1/(1+x^2); y=0 ` if `x=1`

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The differential equations, find a particular solution satisfying the given condition: (1+x^2)(dy)/(dx)+2x y=1/(1+x^2); y=0 when x = 1

The differential equations, find a particular solution satisfying the given condition: (1+x^2)(dy)/(dx)+2x y=1/(1+x^2); y=0 when x = 1

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(1 + x^(2))(dy)/(dx) + 2xy = (1)/(1 + x^(2)), y = 0 when x= 1

(1 + x^(2))(dy)/(dx) + 2xy = (1)/(1 + x^(2)), y = 0 when x= 1

(dy)/(dx)=(y^(2)+y+1)/(x^(2)+x+1)

(dy)/(dx)=(y^(2)-y+1)/(x^(2)-x+1)

The solution of the differential equation x(dy)/(dx) + 2y = x^2 (X ne 0) with y(1) =1, is :