Solving a Homogeneous Differential Equation

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[kingbeauregard]

Note, the quick test for a homogenous equation is if you can replace "x" with "cx" and "y" with "cy" everywhere, and all the c's cancel out. If so, then you can do the "u = y/x" substitution.

[Qermaq]

We were in a good mood, eh? Very fun.

[MrLidless]

I would have used a different substitution, though it’s no better. Call it the second method.

Rearranging the equation, you get y dy/dx - y²/x = x
This highly suggests a u = y² substitution, which gives
du/dx - 2u / x = 2x

Using the general solution to the formula: du/dx +p(x)u = q(x) using integrating factors, we immediately arrive at:
u / x² = y² / x² = integral( x / x²) = integral (1 / x) = ln(x) +c, the same as yours after rearranging.

[moeberry]

The best way to visualize the substitution is to perform the division first and noting that x and y are both nonzero.

[skwbusaidi]

First thing we do
Dy/dx = x^2/(xy) + y^2 /(xy)
= x/y + y/x
Now it is clear to sub u=y/x

[skylardeslypere]

You can simplify the end result by noting that 2ln(|x|) = ln(x²) for all x, and ln(x²)+k = ln(Kx²) where K = e^k

So the final result is ± sqrt(ln(Cx²)) where C is a positive (nonzero) constant.

[florentan]

Other method:

We have: f'(x)=(x^2+f(x)^2)/(xf(x)), then f(x)f'(x)=(x^2+f(x)^2)/x, so [f(x)^2]'/2=(x^2+f(x)^2)/x

If we set g by: g(x):=f(x)^2, then g'(x)/2=(x^2+g(x))/x, so xg'(x)-2g(x)=2x^2.

We find that g(x)=x^2ln(ax^2), and thanks to this, we find f.

[henrymarkson]

By coincidence, earlier today Tom Crawford published a very good YouTube video on such homogeneous first order ODE's.

[jonathanward]

A little Rhyme and Reason
Ol' King Cole was a merry old soul and a merry old soul was he. He called for his pipe and he called for his bowl and he called for his Fiddlers Three.
Ol' Jon Ward was a merry old soul and a merry old soul was he. He called for his pipe and he called for his bowl and he called for his Led Zeppelin CDs located on the right side of the bookcase.
Wanna Whole Lotta Love

y = ux? A little rhyme and reason would have been nice.

[alextang]

It is very wise to use substitution y=ux. But, how can you think out this substitution? Are there any hints to figure out from the right hand side [(x^2)+y^2/xy] of the equation? 🤔🤔🤔

[aboamerometwally]

I solve it by another way
Y can solve it as Nonexact-De solution : I can’t explain the method here but you can use google and write de no exact solution
1- convert it as zero equation
2- multiply both sides by x^ -3
3 go on ...

[thexavier]

"k is for konstant"

Me: the mathematician and the grammar in my soul is in conflict

[shubhamkochar]

You should put some tough questions and some geometrical problems too

[elias]

why use c or k for constant when you can use d for donstant

[abdulhusseinalsultani]

Hello
I have new method to solve some problems of the homogeneous D.E (Alsultani Method )
For the example
dy/dx =(x^2+y^2)/xy
Solution
(X^2+y^2)dx=xydy
(x^2+y^2)dx_xydy=0
The degree of 2
Multipy by 2
2(x^2+y^2)dx_2xydy=0
Alsultani Method
By taking 2y^2dx and _2xydy we get
dy/dx +2y^2/_2xy gives us
dy/dx +2y/_2x so
f(x, y)=y^2/x^2
partial df/partial dy =2y/x^2
2y/x^2=meu(2xy)
Then meu =1/x^3
F(x, y)=f(x, y)+int.(2x^2)dx/x^3
F(x, y)=y^2/x^2+2ln|x|+c
But my method cannot solve the problems like
(x^2+y^2_xy)dx _xydy=0
because of the _xydx and _xydy together.
Also I have another new way to solve the problem (Alsultani D.E instead of Bernoulli D.E)
Solution
dy/dx =(x^2+y^2)/xy=
x/y+y/x
dy/dx _y/x=x/y
dy/dx +P(x)y/n =Q(x)y^k where k=1_n (Alsultani D.E)
dy/dx _y/x=x/y =xy^_1
So k=_1=1_n then n=2
Multiply by 2
2dy/dx _2y/x=2x/y
I.f=e^int.P(x)dx =
e^int._2dx/x=1/x^2 (this is a 2nd I.f.)
So y^2/x^2=2int.xdx/x^2 and
y^2/x^2=2ln|x|+c
Thank you very much

[omograbi]

I tried solving it with triangulometric substitution, but I was lost and got a different solution. Can you make another vid and solve it with triangulometric substitution?

[ChazCharlie]

You could explain d/dx x^2 using the chain rule.
d/dx x^2 = x.x
= x.1 + 1.x
=2×

[chandradeepraut]

Very precisely told, ur videos are always like this with a tone of functional eqns keep it up 🎉🎉🎉

[joannak]

Y? Because we like u! Loved this video and your humor. Keep it up!!

[SpaceTivi]

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