# solving the differential equation y”+t(y’)^2=0

I need to solve:

y″+t(y′)2=0y”+t(y’)^2=0

I did:

u=y′⟹u′=y″u = y’\implies u’= y”

u′+tu2=0⟹u′=−tu2⟹u′u2=−t⟹u’+tu^2=0\implies u’ = -tu^2 \implies \frac{u’}{u^2} = -t\implies
−1u=−t22+c1⟹y′=u=2c1+t2⟹-\frac{1}{u} = -\frac{t^2}{2}+c_1\implies y’ = u = \frac{2}{c_1+t^2}\implies

y=∫2c1+t2dt=2arctant√c1√c1+c2y = \int \frac{2}{c_1+t^2} dt = \frac{2\arctan \frac{t}{\sqrt{c_1}}}{\sqrt{c_1}}+c_2

However, by book presents the following answer:

What am I doing wrong?

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3

You are right if c1>0.c_1>0. But for c1≤0c_1\le 0 the final integral is not an arctan.\arctan.
– mfl
Oct 20 at 21:35

If c1<0c_1<0 you can decompose the denominator – zar Oct 20 at 22:54 ================= =================