Question

Differential equation help *question attached below* will give medal

Answer 1

Usually, I know how to do this. But this is the first time I'm seeing a substitution where dy/dx is involved. Can I have some assistance with this, please?

Answer 2

maybe this identity can be useful:

Answer 3

oops... my image file is not visible

Answer 4

ok! now it can be visualized

Answer 5

I rewrite it:

Answer 6

Hmm, how exactly will it help us in solving the DE?

Answer 7

Using that identity, we can rewrite your ODE as follows:

Answer 8

oops.. I have made a typo error:

Answer 9

that last equation is difficult to solve

Answer 10

Hmm, I'm not getting it :S

Answer 11

I'm curious what's meant by \(\dfrac{dy}{dx}=p\)... is that a constant? Or another function?

Answer 12

@SithsAndGiggles I'm not sure what p is. But since they didn't specify it as a constant, I'm guessing it's another function

Answer 13

using your substitution, we can rewrite your equation as follows:
yp' = p^2
furthermore, using the chain rule we can write:
dp/dx = dp/dy * dy/dx
or:
dp/dx= dp/dy * p
and finally, we get:
p' = dp/dy * p
now using the rewritten ODE, we get:
p^2/ y = p* (dp/dy)
or:
p/y = dp/dy

Answer 14

Okay, so \(y'=p\) is intended to be a hint of some sort.

Answer 15

yes, I think so! @SithsAndGiggles

Answer 16

Let me just process everything

Answer 17

Is this solved? Or do I need to do more? @SithsAndGiggles

Answer 18

Well you have yet to solve for \(p\) (since the ODE is now considering \(p\) as a dependent variable, a function of the independent variable \(y\)). This shouldn't be too difficult seeing as it's separable.

Once you find a solution for \(p\) (i.e. \(p=\cdots\), some function of \(y\)), you would undo the substitution, setting \(y'=p\). Then you have another ODE to solve, this time for \(y\) which depends on \(x\).

Answer 19

I considered \(p = cy^n\). Differentiate that and we get: \[
\frac{d^2y}{dx^2} = \frac{dp}{dx} = \frac{d}{dx}cy^n = cny^ny' =c^2ny^{2n}
\]Then we have: \[
y\frac{d^2y}{dx} = c^2ny^{2n+1}
\]And \[
p^2 = c^2y^{2n}
\]The differential equation ends up as: \[
y^{2n+1}= ny^{2n}
\]I guess this means that power equations are out. I wonder about trig equations.

Answer 20

Okay, here is something else I tried, so we have: \[
yy''=y'y'
\]Divide both sides by \(y\) and \(y'\) and get:\[
\frac{y''}{y'} = \frac{y'}{y}
\]Integrate both sides with respect to \(x\) and you should get: \[
\ln(y')= \ln(y)+C \implies y' = cy
\]I have no idea where \(p\) comes in though.
This looks like a very simple \(y=ke^{cx}\) equation.

Answer 21

If we solve the previous equation, namely:
dp/p = dy/y,
we get:
p=cy, where c is an arbitrary constant
Then we can use the definition of the function p, and so we can write:
p=dy/dx= cy
whose general integral is:
y(x) = k exp(x), where k is another real arbitrary constant

Answer 22

more precisely, the general integral of your ODE is:
\[y(x)=c _{1}\exp(c _{2}x)\]
where c_1 and c_2 are arbitrary real constants.