Find the solution of the given initial value problem in explicit form.

$y\prime=(1-2x)y^2~,~y(0)=\frac{-1}{6}$

Question

Find the solution of the given initial value problem in explicit form.

$y\prime=(1-2x)y^2~,~y(0)=\frac{-1}{6}$

austinl · Answer

I am not even sure where to begin, it isn't in any form that I am familiar with.

loser66 · Answer

$$\frac{dy}{y^2}= (1-2x)dx$$integral both sides

austinl · Answer

$\large\int \dfrac{dy}{y^2}=?$

What? How the...... I feel silly now, but...

loser66 · Answer

yup, $$\int y^{-2}dy=?$$

austinl · Answer

Aha, Now I see. Yep, I feel silly.

loser66 · Answer

ok,

loser66 · Answer

@CarlosGP  watch him out, please. I have to go somewhere to eat something. Cannot stay

austinl · Answer

$\large{-\frac{1}{y}+C=x-x^2+C}$

This time I actually need to solve for y, yes?

anonymous · Answer

If you put C on both sides they will cancel each other:. Put just one C and calculate it using y(0)=-1/6

austinl · Answer

$\large{y(x)=\frac{1}{x(x-1)}+C}$

anonymous · Answer

either way is ok. leaving y(x) alone is not always possible

austinl · Answer

That can't be right.

anonymous · Answer

now do x=0 and get the "C" that makes y(0)=-1/6

austinl · Answer

But... the denominator can't be zero. I am lost now.

anonymous · Answer

.Because the solution is:
$$y(x)=\frac{ 1 }{ x(x-1)+C }$$

anonymous · Answer

It is what you get from:
$$-\frac{ 1 }{ y }=x-x^2+C$$

austinl · Answer

$\large{y(x)=\dfrac{1}{x(x-1)-6}}$

Is this correct?

anonymous · Answer

that is ok. Now check that$$y'=(1-2x)y^2$$

austinl · Answer

Actually,

$\large{y(x)=\dfrac{1}{x^2-x-6}}$

And that is the solution that my text book gives! :D

Thanks very much guys!

anonymous · Answer

you are welcome