Differential Equation - 1st Order: Reducible to Separable Forms (2 of 7) Example 1: 2xyy'-y^2+ x^2=0

In this video I will solve 2xyy'-y^2+ x^2=0.

Next video in the 1st Order: Reducible to Separable Forms series can be seen at:

[Данила-бя]

a wonderful kind man, just a kind man

[doaa]

Thanks for great explanation.. It was really helpul to me!

[nirdeshkunwar]

In 3:38 2u/u^2+1 becomes ln(u^2+1) and would you mind explaining why '2u' was not accounted?

[harshaniperera]

This is sooo amazing😊😊😊😊thank u sir for sharing your knowledge with us🌟🌟🌟

[wisamalkhoory]

Thank you professor, I have a question: is it corret to devide by x sqr since it can be zero and that will make both sides undefined

[iWonderOfficial]

At 3:46, I am confused about what he says about deriving the function. "The derivative of the denominator is 2u du, we have a 2u du, so we can integrate that." I think this is substitution method, but I don't really understand why having the derivative of the denominator in the numerator allows us to integrate the function. Could someone please elaborate? :-)

[ItzYuuri]

Thank you very much for the solution although I have a question and it is, why is the "e" used after everything is in logarithms?

[Ntando]

Prof, it gives me different answer when i take negative to the other side (-In(u^2 + 1)

[jeffersonkho]

Sir, may i ask? Why is that the result becomes different when i put my constant of integration on the equation ln |u²+1|? The final answer becomes x/c=y²+x². Sir, did i make a mistake or violate some rules?

[danielmoss]

I'm still confused. Not by the methodology but by what to use for the substitution. I have spent hours on a question now and I still can't get the right answer :(

[VidyutSquad]

How will be the graphs of solution look like for this?

[AhmetOmerOzgen]

is the e to the power -lnx is = 1/x ? this confused me. thx

[nikulp]

From the second line:
d/dx (y^2/x) = -1
y^2/x = -x + c
y^2 + x^2 = cx.

It's way less work this way, but good video though.

[vibhutigupta]

haven't started with the video but wanted to say your bow tie is really cute!

[younique]

Professor, the way of derivative at 1:40 is because both u and x are functions, not constant?

[abdulhusseinalsultani]

Please I put a new D.E instead of Bernoulli D.E
dy/dx +P(x)y/n =Q(x)y^k where k=1_n (Alsultani D.E )
2xyy^'_ y^2+x^2=0
Solution
Divide by 2xy
y^'_ y/2x= _x/2y= _(xy^_1)/2
k= _1=1_n so n=2
Multiply by 2
2y^' _ y/x =_xy^_1
I.f. =e^int. _dx/x =1/x
y^2/x=_ int.xdx/x= _x+c
Multipy by x
y^2= _x^2+cx
y^2+x^2=cx
Less steps so little time and papers
Thank you very much

[kaylabeckles]

I'm sorry but I think this method is very long and there is an easier way to solve this eqn. begin by carrying your y^2 and x^2 terms to the right then divide throughout by 2xy and you'll probably get a different solution.

[anithareddy]

Is this parabola or hyperbola or ellipses

[bazwaa]

Some mistake in 2:48. 2u^2 - u^2 should be equal to u^2 on the left side, or -u^2 on the right side

[yvonnecheah]

Is that y^2+x^2=cx the final answer of this question?