Question

de

Answer 1

\[yy''+2(y')^2=0\]
Suppose we let \(u=yy'\), then \(u'=(y')^2+yy''\), but then we're still left with a \((y')^2\) term...

Answer 2

Another attempt: Multiply both sides by \(y\), then substitute.
\[y^2y''+2y(y')^2=0\]
Let's try \(u=y^2y'\), then \(u'=2y(y')^2+y^2y''\). Now you have
\[u'=0~~\implies~~u=C~~\implies~~y^2y'=C\]
which is a separable first-order.

Answer 3

yes its separable..

Answer 4

In second order, there are 3 cases right?

Answer 5

Hmm, not sure what you mean by cases. Integrating the final DE gives the exact same answer as you've provided.

Answer 6

Oh.. Thank you so so much =)

Answer 7

Ahm, What are your Techniques in answering second order easily?

Answer 8

Well, admittedly, for this equation I just stared at it for a while :P

Answer 9

As a general technique, I suggest trying a substitution, and if it doesn't work (but gives you something close, as in this case), try modifying the sub or the original equation.

Answer 10

Woah. I got it. Thank you so much.

Answer 11

I think \(u=y''\) substitution always gives you a first order equation

Answer 12

*system

Answer 13

\[yy'' + 2(y')^2 = 0\]

sub \(u(y) = y'(x) \implies u'(y) = y''(x) \dfrac{dx}{dy} \implies y''(x) = u'(y)\dfrac{dy}{dx} = u'(y)u(y) \)

then the equation becomes
\[yu'u + 2^2 = 0\]

which is separable...

looks this substitution works in general and always there to rely on when other tricks wont work...

Answer 14

*
\[yu'u + 2\color{Red}{u}^2 = 0 \]

Answer 15

Thank you so much @ganeshie8 =)

Answer 16

@ganeshie8 hi. Im sorry. Imma little bit confused. How am i gonna reduce this to get the answer? the answer is y^3 = c1x + c2. Thank you

Answer 17

\[yy''+2(y')^2=0~~\iff~~y\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2=0\]
Setting \(u(y)=\dfrac{dy}{dx}\), you have \(\dfrac{du}{dy}=\dfrac{d^2y}{dx^2}\dfrac{dx}{dy}\) by the chain rule. Solving for \(y\)'s second derivative gives \(\dfrac{d^2y}{dx^2}=\dfrac{du}{dy}\dfrac{1}{\frac{dx}{dy}}=\dfrac{du}{dy}\dfrac{dy}{dx}=u\dfrac{du}{dy}\).

Plugging everything relevant into the equation, you get
\[y\left(\frac{du}{dy}\frac{dx}{dy}\right)+2u^2=0~~\implies~~y\left(u\frac{du}{dy}\right)+2u^2=0\]
Separating your variables, you find that
\[\frac{du}{2u}=-\frac{dy}{y}\]
Integrating yields
\[\frac{1}{2}\ln|u|=-\ln|y|+C_1~~\iff~~\sqrt u=\frac{C_1}{y}~~\implies~~u=\frac{C_1}{y^2}\]
Replacing \(u=\dfrac{dy}{dx}\), you have
\[\frac{dy}{dx}=\frac{C_1}{y^2}~~\implies~~y^2\,dy=C_1\,dx~~\implies~~\frac{y^3}{3}=C_1x+C_2\]
as desired.

Answer 18

Thank you so much guys =)