Weird 2nd Order difeq question. Help Please!!

Phyxius117 · #1 $Old$ November 1st, 2009, 01:28 PM

Here is the question:

y'' = (y')^2 (Assume y and y' are positive)

I looked at the question and didn't know how to start it, never seen a question in this form. Help please!!!

Danny · #2 $Old$ November 1st, 2009, 01:57 PM

Quote:

Originally Posted by Phyxius117 $View Post$

Here is the question:

y'' = (y')^2 (Assume y and y' are positive)

I looked at the question and didn't know how to start it, never seen a question in this form. Help please!!!

Divide by $y'$ and integrate once

$\frac{y''}{y'} = y'$ so $\ln y' = y + \ln c_1 \; \text{or}\; y' = c_1 e^y.$

Then separate and integrate again.

Phyxius117 · #3 $Old$ November 1st, 2009, 02:06 PM

Thanks!!

Now I got

y = C1 e^(y) X + C2

as answer but how do I separate to just y in the equation with e^(y) present?

Phyxius117 · #4 $Old$ November 1st, 2009, 02:22 PM

lol I think it's wrong but I plug

y = lny' - C1 into y

and becomes

y= C1 e^(lny' - C1) + C2

which simplifies into

y = C1y' + C2

then I integrate again becomes

yx = C1y + C2x + C3

and finally

y = (C2x + C3) / (X - C1)

which I feel it's wrong...

Danny · #5 $Old$ November 1st, 2009, 03:06 PM

Quote:

Originally Posted by Phyxius117 $View Post$

lol I think it's wrong but I plug

y = lny' - C1 into y

and becomes

y= C1 e^(lny' - C1) + C2

which simplifies into

y = C1y' + C2

then I integrate again becomes

yx = C1y + C2x + C3

and finally

y = (C2x + C3) / (X - C1)

which I feel it's wrong...

Ya - you can't do that!

Quote:

Originally Posted by Danny $View Post$

Divide by $y'$ and integrate once

$\frac{y''}{y'} = y'$ so $\ln y' = y + \ln c_1 \; \text{or}\; y' = c_1 e^y.$

Then separate and integrate again.

From what I have.

$e^{-y} dy = c_1 dx$ so $-e^{-y} = c_1 x + c_2$ (I absorbed the negative into the constants).

Solving for y gives y $= - \ln | c_1 x + c_2|$ .

umgambit · #6 $Old$ November 1st, 2009, 03:22 PM

Just put
y'=f.

The equation then becomes
$\frac{df}{dx}=f^2(x)$

Separate and integrate to get

$f^{-2}df=dx$
and
$-f^{-1}=x+a$ ,
Hence
$f(x)=-\frac{1}{x+a}.$

Now,
$y'=-\frac{1}{x+a}$

Integrate to get the final answer
$y(x)=-\ln(|x+a|)+b$

Phyxius117 · #7 $Old$ November 1st, 2009, 03:29 PM

ty so much!!!

umgambit · #8 $Old$ November 1st, 2009, 03:40 PM

Note that Danny's answer is also correct:

y

$y=-\ln|c_1x+c_2|=-\ln(|c_1(x+\frac {c_2}{c_1}|)=-\ln(|c_1|)-\ln(|x+\frac{c_2}{c_1}|)=-\ln(|x+a|)+b$
where b=-ln(c1) and a=c2/c1