Question #4a5e2

3 Answers
Apr 18, 2017

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)

Apr 18, 2017

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

Apr 18, 2017

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)