# Solve the differential equation dy/dx=8x^2y^2 with the condition given that y(2)=3 .

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We have given

`(dy)/(dx)=8x^2y^2`

In the problem ,variables are separable. Thus

`(dy)/y^2=8x^2dx`

Integrating, we have

`int(dy)/y^2=8intx^2dx+c`

`y^(-2+1)/(-1)=8x^(2+1)/3+c`

`-y^(-1)=(8/3)x^3+c`

Given x=2,y=3 ,thus

`-3^(-1)=(8/3)(2)^3+c`

`c=64/3+1/3=65/3`

`y^(-1)=-{65/3+(8/3)x^3)`

`y=-1/{(65/3)+(8/3)x^3}`

dy/dx=16xy^2/8x^2y, but now with the condition that y(2)=3 what would be the answer then?