hey,

I have this question I have been trying to solve, I got some help and I think I may have solved it but I just wanted to double check here ,

The question is, Consider the non-linear equation below, use y'=p to reduce it to first order, y(o)=1 and y'(0)=-1

y''+y(y')^{3}=0

$\displaystyle p=\frac{dy}{dx}= \pm\frac{1}{\sqrt{y^2- 2c}}$

so

$\displaystyle \pm dy \sqrt{y^2- 2c}= dx$

Then

$\displaystyle y'(0)=\pm \frac{1}{\sqrt{{{y}^{2}}(0)-2c}}=\pm \frac{1}{\sqrt{1-2c}}=-1$

$\displaystyle \pm \frac{1}{\sqrt{1-2c}}=-1,\,\,\,\pm 1=-\sqrt{1-2c}\,\,squaring\,\,both\,\,sides$

1=1-2c

c=0

$\displaystyle So\,y'(x)=\pm \frac{1}{y}$

$\displaystyle \pm ydy=dx$

$\displaystyle \pm \frac{1}{2}{{y}^{2}}=x+c$

$\displaystyle \pm \frac{1}{2}{{y}^{2}}(0)=\pm \frac{1}{2}=c$

$\displaystyle \therefore y(x)=\sqrt{2x\pm \frac{1}{2}}$

Does this look correct?

Thanks in advance