I'm sure it's staring at me, but how does one solve this?
(y')^2 + y = xy'
Thanks.
I'm sure it's staring at me, but how does one solve this?
(y')^2 + y = xy'
Thanks.
Differentiating both sides with respect to x:
$2y^\prime y^{\prime\prime}+y^\prime =y^\prime+xy^{\prime\prime}$
i.e., $y^{\prime\prime}(2y^\prime −x)=0$ which can now just be solved for each case in turn, $y^{\prime\prime}=0$ and $2y^\prime=x$
Thanks everyone.