Question

Solve 2nd Order DE\[y\frac{\text d^2y}{\text dx^2}+1=\left(\frac{\text dy}{\text dx}\right)^2\]

Answer 1

http://www4c.wolframalpha.com/Calculate/MSP/MSP2831a22ehagbf6fe27h00005c3g3ihb628g7697?MSPStoreType=image/gif&s=37&w=431&h=1458

Answer 2

Interesting method ...
$$
y \frac{dv}{dy}v + 1 = v^2
$$

Answer 3

let dy/dx = v

Answer 4

let dy/dx=p
then d^2y/dx^2=dp/dx=(dp/dy)(dy/dx)=p(dp/dy)

Answer 5

then we have

Answer 6

\[yp(dp/dy)+1=p^2\]

Answer 7

\[y\frac{\text d^2y}{\text dx^2}+1=\left(\frac{\text dy}{\text dx}\right)^2\]\[\text {let } p=\frac{\text dy}{\text dx}\qquad\qquad\frac{\text dp}{\text dx}=\frac{\text d^2y}{\text dx^2}=p\frac{\text dp}{\text dy}\]\[yp \frac{\text dp}{\text dy}+1=p^2\]

Answer 8

\[ \frac{dp}{dx} = \frac{dp}{dy} \frac{dy}{dx} = \frac{dp}{dy} p\]

Answer 9

\[yp \frac{\text dp}{\text dy}+1=p^2\]\[yp \frac{\text dp}{\text dy}=p^2-1\]\[\frac{p\text dp}{p^2-1}=\frac{\text dy}{y}\]\[\frac 12 \int\frac{2p}{p^2-1}\cdot \text dp=\int\frac{\text dy}{y}\]

Answer 10

i made a terrible mistake up there...lol

Answer 11

\[\frac 12\ln|p^2-1|=\ln|cy|\]\[p^2-1=Ay^2\]\[\left(\frac{\text dy}{\text dx}\right)^2-1=Ay^2\]

Answer 12

a) have i done this right sofar,
b) what can i do next

Answer 13

I'm not sure if you can do this but, what if you take roots at both sides and; (y')^2=Ay^2 + 1; y'=sqrt(Ay^2+1); separate variables and you have; dy/sqrt(Ay^2+1)=dx; 1/2*argsinh(y)=x + C ???

Answer 14

@UnkleRhaukus your steps seem correct.
Now, \[(\frac{dy}{dx})^{2} = (Ay^{2}+1)\]\[\frac{dy}{dx} = \pm\sqrt{1+Ay^{2}}\]
\[\int\limits \frac{dy}{\sqrt{1+Ay^{2}}} = \int\limits \pm dx\]Substitute
\[y = \tan\theta\sqrt{A}\]
See if you can take it from here.

Answer 15

I think the sub would be\[y=\frac1{\sqrt A}\tan\theta\]

Answer 16

...but is that \(\pm\) a wrench in the works?
of that I'm not sure

Answer 17

the answer in the back of the book has logs rather than arctans

Answer 18

so i think i hav to use this integral for the table
\[[20]\qquad\int\frac{\text dt}{a^2-x^2}=\frac{1}{2a}\ln\left|\frac{x+a}{x-a}\right|+c\]
somehow

Answer 19

\[p^2-1=c^2y^2\]