Question

Here is an interesting question.

Find \(y=f(x)\):\[
\frac{d}{dy}\frac{dy}{dx} = \frac{5}{x}
\]Since it's a differential equation, there will be a general solution, but I don't necessarily want that.
I just want to see how people approach this problem.

Answer 1

How I got here:
I was thinking about how order of differentiation matters. \[
\frac{d}{dx}\frac{dy}{dy} = \frac{d}{dx}1 = 0
\]But that isn't necessarily the case for \(\frac{d}{dy}\frac{dy}{dx}\).

Answer 2

In essence: \[
\frac{d}{dy}\frac{d}{dx} u \neq \frac{d}{dx}\frac{d}{dy} u
\]

Answer 3

Oh you editted it, just noticed.

Answer 4

Yeah, there was a latex mistake.

Answer 5

implicitly differentiating \(\dfrac{5}{x}\) would have you ending up with \(5\ln(x)\) somehow, right? Just a guess. x_x

Answer 6

What are you guessing about?

Answer 7

just a guess \[y = cx^5\]

Answer 8

Close, but not quite. How did you get that anyway?

Answer 9

try this, \[y=x^5\implies y' = \frac{5y}{x} \implies \frac{d}{dy} y' = \frac{5}{x}\]

Answer 10

it was just a blind guess..

Answer 11

I am trying to legitimately solve this. I set up \(y=f(x)\) before hand and then differentiated.

Answer 12

You can get it into the form: \[
xy''-5y'=0
\]

Answer 13

The say \(u=y'\) and: \[
xu'-5u=0
\]

Answer 14

So \[
u'-\frac 5xu = 0
\]

Answer 15

try this, \[y=x^5\implies y' = \frac{5y}{x} \implies \frac{d}{dy} y' = \frac{5}{x}\]The problem with this is the fact that \(x\) is not constant with respect to \(y\).

Answer 16

why are you posting pre-university problems

Answer 17

oops... the denominator is:
\[\ln(e/(x ^{5}))\]

Answer 18

Again, you treat \(x\) as constant with respect to \(y\). When you integrate \(x\) with respect to \(y\), you are essentially integrating \(f^{-1}\) since it is the inverse.

Answer 19

I'm sorry, my answer is wrong, please neglect it!

Answer 20

If I integrate once your equation, I can write as more general solution, the function below:
\[\frac{ dy }{ dx }=\frac{ 5 }{ x }y+g(x)+c\]
where c is first integration constant, and g(x) is a function of x only.
Now I can rewrite that equation as below:
\[y'-\frac{ 5 }{ x }y=g(x)\]
which can easily integrated with standard methods, and whose solution is:
\[y=f(x)=x ^{5}\left( c+\int\limits \frac{ g(x) }{ x ^{5} } \right)\],
where c is the second integration constant

Answer 21

sorry there are some errors in integration constants:
\[\frac{ dy }{ dx }=\frac{ 5 }{ x }y+g(x)+c _{1}\]
so:
\[y'-\frac{ 5 }{ x }y=g(x)+c _{1}\]
and:
\[y=f(x)=x ^{5}\left( c _{2}+\int\limits \frac{ g(x)+c _{1} }{ x ^{5} } dx\right)\]
where c_1 and c_2 are the constants of integration

Answer 22

please note that if I calculate the first derivative with respect to y of the subsequent function:
\[\frac{ d y}{ dx }=\frac{ 5 }{ x }y+g(x)+c _{1}\]
I will get:
\[\frac{ d }{ dy }\frac{ dy }{ dx }=\frac{ 5 }{ x }\]

Answer 23

Okay, but I'll say it again... \[
\int\frac 5xdy \neq \frac 5x\int dy
\]

Answer 24

Since \(y=f(x)\), that means \(x = f^{-1}(y)\).

Answer 25

Which is why \(5/f^{-1}(y)\) can't be pulled out.

Answer 26

Random thought, I see we can get to:\[\frac{dy}{dx}=\frac{5y}{x}+C\]How about then converting this to polar coordinates since the \(\frac{y}{x}\) term will then simplify to \(\tan(\theta)\)?

Answer 27

\[
\frac{dy}{dx}\neq \frac 5xy+C
\]

Answer 28

Actually, it might in this case, but not generally I don't believe.

Answer 29

Consider if \(y=\sin(x)\).\[
\frac{dy}{dx} = \cos(x) = \sqrt{1-\sin^2(x)} = \sqrt{1-y^2}
\]And \[
\frac{d}{dy} \frac{dy}{dx} =\frac{y}{\sqrt{1-y^2}} = \tan(x)
\]Would you say that: \[
\frac{d}{dx}\sin(x) = \tan(x)\sin(x)
\]?

Answer 30

I'll give this hint, and go to bed:
Let \(u=dy/dx\)\[
\frac{du}{dy} = \frac 5x
\]Multiply both sides by \(dy/dx\):\[
\frac{du}{dy}\frac{dy}{dx} = \frac{5}{x}\frac{dy}{dx}
\]

Answer 31

And for a bonus, see if you guys can figure out \(y\), such that:\[
\frac{d}{dy}\frac{dy}{dx} = y
\]

Answer 32

\[\frac{du}{dy}\frac{dy}{dx} = \frac{5}{x}\frac{dy}{dx}\]\[\therefore \frac{du}{dx}=\frac{5}{x}\frac{dy}{dx}=\frac{5u}{x}\]\[\therefore\int\frac{1}{u}du=\int\frac{5}{x}dx\]\[\therefore\log(u)=5\log(x)+C=\log(x^5)+\log(k)=\log(kx^5)\]\[\therefore u=kx^5\]\[\therefore \frac{dy}{dx}=kx^5\]etc...

Answer 33

For the other one, again let:\[u=\frac{dy}{dx}\]\[\therefore \frac{du}{dy}=y\]\[\therefore \int du=\int ydy\]\[\therefore u=\frac{y^2}{2}+C=\frac{y^2+a^2}{2}\]\[\therefore\frac{dy}{dx}=\frac{y^2+a^2}{2}\]\[\therefore \int\frac{1}{y^2+a^2}dy=\frac{1}{2}\int dx\]\[\therefore\frac{\tan^{-1}(\frac{y}{a})}{a}=\frac{x}{2}+C\]\[\therefore \tan^{-1}(\frac{y}{a})=\frac{ax}{2}+D\]\[\therefore \frac{y}{a}=\tan(\frac{ax}{2}+D)\]\[\therefore y=a\tan(\frac{ax}{2}+D)\]