What is the (dy)/(dx)=(1+3x^2)/((1+y)^2),y(1)=2 ?

The (dy)/(dx)=(1+3x^2)/((1+y)^2),y(1)=2 is 1/9+(x^2)/3

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(dy)/(dx)=(1+3x^2)/((1+y)^2),y(1)=2

\frac{d y}{d x} = \frac{1 + 3 x ^{2}}{( 1 + y ) ^{2}}, y (1) = 2

\frac{1}{9} + \frac{x ^{2}}{3}

Solution steps

\frac{1 + 3 x ^{2}}{( 1 + y ) ^{2}} = \frac{1 + 3 x ^{2}}{( 1 + y ) ^{2}}

Constant function results in a constant

y (1) = 2

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

Expand

\frac{1 + 3 x ^{2}}{( 1 + 2 ) ^{2}} : \frac{1}{9} + \frac{x ^{2}}{3}

= \frac{1}{9} + \frac{x ^{2}}{3}

What is the (dy)/(dx)=(1+3x^2)/((1+y)^2),y(1)=2 ?
The (dy)/(dx)=(1+3x^2)/((1+y)^2),y(1)=2 is 1/9+(x^2)/3