1. ## Weird 2nd Order difeq question. Help Please!!

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!!!

2. Originally Posted by Phyxius117
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 $\displaystyle y'$ and integrate once

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

Then separate and integrate again.

3. 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?

4. 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...

5. Originally Posted by Phyxius117
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!

Originally Posted by Danny
Divide by $\displaystyle y'$ and integrate once

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

Then separate and integrate again.
From what I have.

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

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

6. Just put
y'=f.

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

Separate and integrate to get

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

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

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

7. ty so much!!!

8. Note that Danny's answer is also correct:

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