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}

\int e^{- s \cdot t} d t

t^{2} y^{^{''}} + 2 t y^{^{'}} - 2 y = 0

\int \frac{4 x + 3}{2 x + 1} d x

y = x^{3} - 16 x + 2

\int r^{- 2} d r

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