Question

Solve the given differential equation by using the substitution \(u=y'\)
\[y''+(y')^2+1=0, u\frac{du}{dy}=y''\]
\[udu+u^2dy+dy=0\]
\[udu=-u^2dy-dy=-dy(u^2+1)\]
\[\int \frac{u}{u^2+1}du=-\int dy\]
\[v=u^2+1,\frac{dv}{2}=udu\]
\[\frac{1}{2}\int \frac{dv}{v}=-\int dy\]
\[\ln(u^2+1)=-2(y+\ln(c_1))\]
\[u^2=c_1^{-2}e^{-2y}-1\]
\[u=\pm\sqrt{c_1^{-2}e^{-2y}-1}=\frac{dy}{dx}\]
I'm not sure where I can go from here.

Answer 1

there is a mistake in the fist line

Answer 2

\(\large y' = u \implies y'' = u'\)

so the equation must become become :

\(\large u' + u^2 = -1\)

Answer 3

\(\large \dfrac{du}{dx}= -(1+u^2)\)

\(\large \int \dfrac{du}{1+u^2} = \int - dx\)

Answer 4

right ?

Answer 5

It's looking like that will get the answer in the book. I did the substitution based on the professor mentioning using that as the chain rule, but there is no y term.

Answer 6

yes this substitution works only when there is no y term

Answer 7

DE needs to be of form
f(y'', y', x) = 0

Answer 8

\[y''+(y')^2+1=0, u'=y''\]
\[u'+u^2+1=0\]
\[\int \frac{du}{u^2+1}=-\int dx\]
\[\frac{1}{2}\int \frac{dv}{v}=-\int dx\]
\[\arctan(u)=c_1-x\]
\[u=\tan(c_1-x)\]
\[dy=\tan(c_1-x)\
\[u=\cos(c_1-x_,du=\sin(c_1-x)dx\]
\[dy=\frac{du}{u}\]
\[y=\ln|cos(c_1-x)|+c_2\]
And that is correct.

Answer 9

looks good to me :)