Solving the Initial-Value Problem: Step-by-Step Guide for Beginners

[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

 

[radou]

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

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.

 

[bobbarkernar]

so am i solving for C?

 

[radou]

so am i solving for C?

Yes, you are.

 

[bobbarkernar]

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

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

 

[radou]

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

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

Yes, C = -16.

 

[bobbarkernar]

negative?

 

[radou]

negative?

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

 

[bobbarkernar]

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

 

[radou]

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

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

 

[bobbarkernar]

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

 

[bobbarkernar]

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

thank you for your help

 

[radou]

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

thank you for your help

Yes, I just wanted to write that the answer is a function, not a constant.