solve the differential equation dy/dx=(8x^2*y^2) with the condition that y(2)=3

the solution to the equation is y=

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Solve `(dy)/(dx)=8x^2y^2` with the initial condition y(2)=3:

This is separable so we get all expressions involving y on one side of the equation, and x on the other:

`(dy)/(y^2)=8x^2dx` Integrate both sides:

`int y^(-2)dy=8int x^2dx`

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

From the initial condition, when x=2 y=3 so:

`-1/3=64/3+C`

`C=-65/3`

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

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

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

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We can check this by differentiation:

Note that `y^2=9/((8x^3-65)^2)`

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

`(dy)/(dx)=(-3(-1)(24x^2))/((8x^3-65)^2)`

`=9/((8x^3-65)^2)*8x^2`

`=8x^2y^2` as required.

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