Question

Q:Solve the 2nd order DE\[y\frac{\text d^2y}{\text dx^2}+1=\left(\frac{\text dy}{\text dx}\right)^2\]
A:\[\pm c\ln(y+\sqrt{y^2+c^2})=x+d\]

Answer 1

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

Answer 2

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

Answer 3

\[[20]\qquad\int\frac{\text dt}{a^2-x^2}=\frac{1}{2a}\ln\left|\frac{x+a}{x-a}\right|+c\]

Answer 4

@UnkleRhaukus
I don't think that your third post is correct. The correct substitution after the second post would be
\[y = \frac{1}{\sqrt{c}}\tan\theta\] which would give a logarithmic answer as the integral comes to be \[y = \int sec\theta d\theta\]

Answer 5

\[p^2-1=c^2y^2\]\[p=\pm\sqrt{c^2y^2+1}\]\[\frac{\text dy}{\text dx}=\pm\sqrt{c^2y^2+1}\]\[\pm\int\frac{\text dy}{\sqrt{c^2y^2+1}}=\int{\text dx}\]

Answer 6

i dont understand the substitution you are suggesting @FoolAroundMath

Answer 7

\[[19]\qquad\int\frac{\text dx}{a^2+x^2}=\frac 1a \arctan \left(\frac xa \right)+c\]

Answer 8

\[\dots?\]

Answer 9

That is what Mathematica gave as a solution
\[
\begin{array}{c}
y(x)=-e^{-c_1} \sinh \left(e^{c_1}
\left(x+c_2\right)\right) \\
y(x)=e^{-c_1} \sinh \left(e^{c_1}
\left(x+c_2\right)\right) \\
\end{array}

\]

Answer 10

i am trying to get to the answer i posed in the question box,
maybe try rearranging before entering into mathematica

Answer 11

@UnkleRhaukus
the formula for integral of 1/(a^2+x^2) doesnt have a square root, whereas the one that we got does.
The required substitution. (edit: I didn't see you had taken the constant to be c^2) would be
y = tan@/c

Answer 12

\[\pm \int\limits \frac{dy}{\sqrt{c^{2}y^{2}+1}} = \int\limits dx\] substitute \[y=\frac{tan\theta}{c}\] so that \[dy = \frac{sec^{2}\theta d\theta}{c}\] and \[\sqrt{1+y^{2}c^{2}} = sec\theta\]
replacing the terms we get
\[\pm \int \frac{sec\theta}{c}d\theta = \int dx\] which leads to
\[\pm \frac{ln(sec\theta+tan\theta)}{c} = x +d \]
now substituting for tan@ and sec@, we get
\[\pm \frac{ln(\sqrt{1+y^{2}c^{2}}+cy)}{c}= x+d\]
Replace 1/c with c' to get what the given answer is.

Answer 13

\[\pm\int\frac{\text dy}{\sqrt{c^2y^2+1}}=\int{\text dx}\]\[\text{let }y=\frac 1c\tan\theta\]\[\text dy=\frac 1c\sec^2\theta\cdot\text d\theta\]\[\pm\int\frac{\frac 1c\sec^2\theta\cdot\text d\theta}{\sqrt{c^2\left(\frac 1c\tan\theta\right)^2+1}}=\int{\text dx}\]\[\pm\frac 1c\int\frac{\sec^2\theta\cdot\text d\theta}{\sqrt{\tan^2\theta+1}}=\int{\text dx}\]\[\pm\frac 1c\int\sec\theta\cdot\text d\theta=\int{\text dx}\]

Answer 14

\[\pm\frac 1c\int \sec \theta \times\frac{\sec \theta+\tan \theta}{\sec \theta+\tan \theta}\text d\theta=\int{\text dx}\]\[\pm\frac 1c\int \frac{\sec^2 \theta+\sec \theta\tan \theta}{\sec \theta+\tan \theta}\text d\theta=\int{\text dx}\]
\[\text{let } u={\sec \theta+\tan \theta}\]\[\text du=\left(\tan \theta\sec x+\sec^2\theta\right)\text d\theta\]
\[\pm\frac 1c\int\frac{\text du}{u}=\int{\text dx}\]\[\pm\frac 1c\ln u=x+d\]\[\pm\frac 1c\ln|\sec \theta+\tan \theta|=x+d\]

Answer 15

now what can i do?

Answer 16

\[tan\theta = cy, sec\theta = \sqrt{1+c^{2}y^{2}}\]Substitute above instead of \[\theta = arctan(cy)\]

Answer 17

how-come ?
\[\sec\theta = \sqrt{1+c^{2}y^{2}}\]

Answer 18

\[\pm\frac 1c\ln|\sec \theta+\tan \theta|=x+d\]\[\pm\frac 1c\ln|\sqrt{ 1+c^2y^2}+cy|=x+d\]

Answer 19

@experimentX

Answer 20

we had substituted \[y=\frac{1}{c}tan\theta, => tan\theta = cy\] \[sec\theta = \sqrt{1+tan^{2}\theta }\]

Answer 21

that makes sense @FoolAroundMath

Answer 22

Wolf gives me
http://www.wolframalpha.com/input/?i=integrate+1+%2Fsqrt%28cx^2%2B1%29

Answer 23

if you are using constant c .. you can drop +-

Answer 24

it seems okay ... i 'm quite comfortable with hyperbolic functions.