Nice tricks to solve this Differential Equation.

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[themanwhocarriesthesun]

Multiply both sides by x, so you have d(xy)/dx = (x^2)*sqrt(y). Let u = xy, so du/dx = x*sqrt(x)*sqrt(u) or du/sqrt(u) = x^(3/2)dx. Integrate both sides, you can find the solution.

[tomatrix]

Never stop making videos Michael, you’re so good at this stuff it’s unbelievable, honestly.

[JM-usfr]

I actually enjoyed ODE’s back when I took it, despite being more algebra and number theory focused.

[goodplacetostop]

6:46 Remember students, exams should be not be viewed as a measure of your worth. You will remain a valuable person regardless of the outcome of any exam. Have a good day!

[jamiewalker]

I made the substitutiion y = u^2 x^4 which turns the equation into a very easy separable differential equation. What inspired me to do this, was to realise that if y has dimensions of some power of x, then the left hand side is automatically dimensionally consistent, and the right hand side could be made consistent with the left, provided that y varies with the fourth power of x. (you can actually see this pop out of your solution when setting c = 0). The u^2 was just to make taking the square root easier.

[manucitomx]

I loved this problem.
Thank you, professor.

[AriosJentu]

Maybe you'll get interested in something like this equation:
(x - cosy) y'' + (siny) * y'^2 + 2y' = 0
It's a specific type equation, which can be integrated twice, which I have derived in my little research work

[nikitakipriyanov]

2:44 z=sqrt(y) will be a standard substitution for a Bernoulli DE like this. We should see this substution from the very beginning just by looking at the DE, without any prior transformations.

Generally, for the equation y'+p(x)y=q(x)y^n we have the substitution z = y^(1-n), which turns the equation into z' + (1-n) p(x) z = (1-n) q(x). The resulting equation for z is a linear first order non-homogeneous and could be solved using integrating function. In this case, we have n=½ p=1/x, q=x.

[Kapomafioso]

(wrote this before watching the video) If you set y = u^2 (motivated by that square root) then:
y' = 2 u u'
so
2 u u' + u^2/x = x u
One of the solutions is y = u = 0. Assuming u is not 0, we can divide by u
2 u' + u/x = x

This is a linear differential equation, so let's use the standard tricks. First, the general solution (zero left-hand side)
2 u' + u/x = 0
2u'/u = -1/x
2 ln u = - ln x + C
u (x) = C/sqrt(x)

Now for the particular solution - the usual trick is to spot some kind of chain rule, which becomes obvious after multiplying by sqrt(x) / 2
sqrt(x) u' + u / (2 sqrt(x)) = (1/2) x^(3/2)

This can be written as
(sqrt(x) u)' = (1/2) x^(3/2)

Since we already captured the homogenous solutions, we don't have to worry about constants here
sqrt(x) u = (1/5) x^(5/2)
u (x) = x^2/5

So the general solution is
u(x) = x^2/5 + C/sqrt(x)

Or in terms of y
y(x) = u(x)^2 = (x^2/5 + C/sqrt(x))^2

I see Michael's solution is basically same.

[holyshit]

Bernoulli equation can be solved in the same way as linear equation
There exists integrating factor in separable form
Variation of parameter also work here

[hamdiel-sissi]

Unnecessarily convoluted solution!!

The original equation can be written as x*dy/dx+y=x**2 *y**1/2
Or d(xy)/dx= x**3/2 *(xy)**1/2

Substitute (xy) with z to get

d(z)/dz=x**3/2*z**1/2 or dz/dx*z**-1/2= x**3/2

which reduces to 2z**1/2=2/5*x**5/2+ C

Or y= (1/5x**2+Cx**-1/2)**2

[jeremyhuang]

For all natural number n greater or equal to 2 : ((n+1)!)^(1/n) - (n!)^(1/n) > 1
By induction or other method

[matematicacommarcospaulo]

I would like to learn how you do to knock the board and something magically apears at 3:23. It is not the 1st time you did this...

[alexpotts]

It's a bit weird, but I've never seen anyone use that differential equation formula, it's just a little bit too clunky to remember and you honestly get a deeper understanding of the topic if you go through the whole throught process - "I want to make the left-hand side the derivative of a product, to do that I will use the integrating factor, etc".

But then I'm a bit weird and I never even use the quadratic formula (even though that's something that nobody ever forgets) because it feels much more like real maths if you complete the square properly.

[mcwulf]

I thought I found a quick method but I get a different answer.

Multiply through by X to get
xy' + y = x^2 ✓y

The LHS is the derivative of xy.
d(xy)/dx

Put z = xy and we get
z' = ✓z . ✓x^3

Divide by ✓z and integrate to get
✓z = (1/5) ✓x^5 + k
after multiplying by 1/2.

Rearrange and put z=xy gives
25y = x^4 + 2k✓x^3 + k^2/x

[MrNygiz]

i multiplied by x and got xy'+y=x^2sqrt(y) => (xy)'=x^2sqrt(y) then set z=xy to get z'=x^3/2sqrt(z) which is seperable, and proceded from there

[pasqueocwe]

Learning matematica wth Michael Penn good

[pikkutonttu]

I guessed that this initially hard looking differential equation is Bernoulli type. However I can't rememeber how to solve it without help. I tried my usual symbolic trickery. Firstly D ln xy = 1/x + y'/y, so we have D ln xy = x/sqrt y. That is at least symbolicly equivalent to INT sqrt xy d ln xy = INT x^3/2 dx. More trickery to the first part and the integral is equal to INT exp (1/2 ln xy) d ln xy = 2exp(1/2 ln xy) = 2sqrt xy. Secondly the right side is equal to 2/5 x^5/2 + c. So we have to solve equation sqrt xy = 1/5 x^5/2 + c', and we get y = (1/5 x^2 + c'/sqrt x)^2. How to justify the symbolic manipulations.

[robberbarron]

Every video starts the same "here we've got a nice..."

[jopetdevera]

Oooh yeah that's pretty neat, though it can be solved right away. This is much cleaner though.