Question

Find the general solution to the following separable DE:
dy/y^2=2xdx/sqrt(1+2x^2)

Answer 1

\[\frac{dy}{y^2}=\frac{2x~dx}{\sqrt{1+2x^2}}\]
Integrating both sides (I'll forgo the constant for now), you get
\[-\frac{1}{y}=2\int\frac{x}{\sqrt{1+2x^2}}~dx\]
For the RHS, I'd suggest a trig substitution: \(x=\dfrac{1}{\sqrt2}\tan u\), so that \(dx=\dfrac{1}{\sqrt2}\sec^2u~du\).

Answer 2

can u show the full solution?? i'm very weak in this topic :((

Answer 3

Not the full, sorry. I've been warned before!

Using the substitution I mentioned earlier, the integral becomes
\[-\frac{1}{y}=2\int \frac{ \frac{1}{\sqrt2}\tan u }{ \sqrt{ 1+2\left(\frac{1}{\sqrt2}\tan u\right)^2 } }~ \left(\frac{1}{\sqrt2}\sec^2u~du\right)\]
Some simplifying gives you
\[-\frac{1}{y}=2\cdot\frac{1}{\sqrt2}\cdot\frac{1}{\sqrt2}\int \frac{ \tan u \sec^2u}{ \sqrt{ 1+2\left(\frac{1}{2}\tan^2 u\right) } }~du\\
-\frac{1}{y}=\int \frac{ \tan u \sec^2u}{ \sqrt{ 1+\tan^2 u } }~du
\]

Answer 4

@chriszeyin, actually, as it's been pointed out to me, the method I'm suggesting is more work! Forget everything I've suggested here.

Instead, use the substitution \(u=1+2x^2\).

Answer 5

ouch sry :( correct me if im wrong..dy/y^2 = 1/y^2 then it become y^-2. after u integrate it..it becomes y^-1 or 1/y..but how u get -1/y? oh ya u=1+2x^2 is better :) thx fr giving the idea

Answer 6

\[\int\frac{dy}{y^2}=\int y^{-2}~dy=\frac{y^{-1}}{-1}+C=-\frac{1}{y}+C\]

Answer 7

ahh!! i forgot the basic of integration :S by the way thx again for helping me :)

Answer 8

You're welcome! My other method works too, but if you haven't learned it yet/are having trouble with it, always go for the easier one

Answer 9

oh ya..i showed the solution to my lecturer and he said is wrong :S
the final answer is [(1+2X^2)^1/2]+1/y=C :((( i have no idea how they get this answer

Answer 10

\[\frac{dy}{y^2}=\frac{2x~dx}{\sqrt{1+2x^2}}\]
Integrating, and letting \(u=1+2x^2~\Rightarrow~du=4x~dx\),
\[-\frac{1}{y}=\frac{1}{2}\int\frac{du}{\sqrt u}\\
-\frac{1}{y}=\frac{1}{2}\left(\frac{u^{1/2}}{\frac{1}{2}}\right)+C\\
-\frac{1}{y}=\sqrt u+C\\
-\frac{1}{y}=\sqrt {1+2x^2}+C\\
C=\sqrt {1+2x^2}+\frac{1}{y}\]
\(C\) is an arbitrary constant, so it doesn't matter which side of the equation you put it.