# Initial-value problem

1. Oct 23, 2006

### bobbarkernar

i have no idea on how to start this problem.

Find a solution of the initial-value problem

dy/dx= -y^2 , y(0)=.0625

so..

(1/-y^2)dy = dx

int(1/-y^2)dy=int(dx)

1/y=x+c
please if someone can help me start it or explain it to me

2. Oct 23, 2006

Note that you have $$-\frac{1}{y} = x + C$$. Now all you have to do is write $$y = -\frac{1}{x+C}$$ and use the initial value.

Last edited: Oct 23, 2006
3. Oct 23, 2006

### bobbarkernar

so am i solving for C?

4. Oct 23, 2006

Yes, you are.

5. Oct 23, 2006

### bobbarkernar

so
C= (1/y)-x ??

so C=(1/.0625) if y(0)=.0625 ????
is this correct?

6. Oct 23, 2006

Yes, C = -16.

7. Oct 23, 2006

### bobbarkernar

negative???

8. Oct 23, 2006

Yes, since $$y(x) = -\frac{1}{x+C} \Rightarrow y(0) = -\frac{1}{C} = 0.0625$$

9. Oct 23, 2006

### bobbarkernar

but the integral of -1/y^2 dy is 1/y

10. Oct 23, 2006

Yes, you're right. I apologize, that minus sign slipped through somehow. So, your answer is correct.

11. Oct 23, 2006

### bobbarkernar

ok but i put the answer c=16 and it was wrong. is there another step?

12. Oct 23, 2006

### bobbarkernar

ok i got it the answe is y=1/(x+16)