Question

Question #4a5e2

Answer 1

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

Explanation:

Given: `dy/dx = xy^2; y(0)=1`

Use the separation of variable method:

`dy/y^2= xdx`

Integrate:

`intdy/y^2= intxdx`

`-1/y = x^2/2+C`

`-2/y = x^2+C`

`y = -2/(x^2+ C)`

Evaluate at the boundary condition, `y(0) = 1`:

`1 = -2/(0^2+C)`

`C =-2`

The equation is:

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

Answer 2

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

Explanation:

`(dy)/(dx) = xy^2`

`1/y^2 dy= x dx`

`int 1/y^2 dy= int x dx`

`-1/y = x^2/2 +c`

plug in `y = 1 and x =0` in the above equation

`-1/1 = 0^2/2 +c` `-> c =-1`

therefore,
`-1/y = x^2/2 -1 = (x^2 - 2)/2`

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

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

Answer 3

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

Explanation:

`dy/dx=xy^2`

`inty^-2` `dy=intx` `dx`

`-1/y=1/2x^2+"c"`

Given that `y=1, x=0`,

`-1/1=1/2(0)^2+"c" rArr"c"=-1`

Since they haven't asked to give it in the form `y=f(x)`, you can leave it like this. But I will solve it for `y` just in case you would like to see how to do that.

`-1/y=1/2x^2-1`

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

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

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