Solve the initial vale problem `(dy)/(dx)+2y^2=0,\ y(1)=1\ ` and find the corresponding solution curve.

Answer

`y=1/(2x-1)`

Explanation

We have given,

`(dy)/(dx)+2y^2=0`


`(dy)/(dx)=-2y^2`


`(dx)/(dy)=-1/(2y^2)`


Separate the variables,

`dx=-1/(2y^2)dy`

integrating both sides, we got,

`int dx=int -1/(2y^2)dy`


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