Question

Can someone solve this...xyy'=1-x^2?....thanks :)

Answer 1

answer
`y'=(1-x^2)/(x*y)`

Explanation:

i think that wanted
`xy*y'=1-x^2`

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

Answer 2

`y=sqrt(2lnx-x^2-c_1)`

Explanation:

First rewrite the differential equation. (Assume `y'` is just `dy/dx`):
`xydy/dx=1-x^2`

Next, separate the x's and y's- just divide both sides by `x` and multiply both sides by `dx` to get:
`ydy=(1-x^2)/xdx`

Now we can integrate both sides and solve for y:
`intydy=int(1-x^2)/xdx`
`intydy=int1/xdx-intx^2/xdx`
`y^2/2+c=lnx-intxdx`
(You only need to put the constant on one side because they cancel each other out into just one `c`.)

(Solving for y):
`y^2/2=lnx-x^2/2-c`
`y^2=2lnx-x^2-c_1` . (Can change to `c_1` after multiplying by 2)
`y=sqrt(2lnx-x^2-c_1)`