Solving A Differential Equation | 2nd method?

[@umut1269]

4:38 sarı laciverti bir arada görmek sevindirdi 😅 as always, great explanation and showing how to approach to a problem. Yours and these kind of channels keeps my passion about math always sharp and alive .

[@SyberMath]

Thanks! 😃

[@leif1075]

​@SyberMath Thanks for sharing Syber. I don't mean to bother you but I hope you can please respond to my question about sum of squares integers formula.

[@cameronspalding9792]

@ 7:32 you need to include the +1 on the numerator

[@holyshit922]

Bernoulli equation
Here exists integrating factor
mu(x,y) = exp((1-r)Int(p(x),x))y^{-r}
but it can be solved by assuming that y(x)=u(x)v(x)
and solving separable equation twice
y'=(x-y^2+1)/(2xy)
y' = -y/(2x) + (x+1)/(2x)*1/y
y'+1/(2x)y = (x+1)/(2x)*1/y
p(x) = 1/(2x)
r = -1
mu(x,y) = exp((1-(-1))Int(1/(2x),x))y^{-(-1)}
mu(x,y) = exp(ln(x))y^{1}
mu(x,y) = xy
xyy'+1/2y^2 = (x-1)/2
d/dx(x*1/2y^2) = d/dx(1/4(x+1)^2)
1/2xy^2 = 1/4(x+1)^2+C_{1}
y^2 = 1/2(x+1)^2/x+C/x

[@florianbuerzle2703]

You could separate the fraction and thus rewrite the DE as
y' + (1 / (2x)) y = (1/2 + 1 / (2x)) y^(-1),
so it has the form
y' + P(x) y = Q(x) y^(-1)
and is therefore a Bernoulli differential equation which can be solved using the substitution u = y², which leads to the equation du/dx + u/x = 1 + 1/x
and multiplying this by x gives
x du/dx + u = x + 1
which is equivalent to
x du + (u - x - 1) dx = 0.
This is an exact differential equation with the solution
ux - x²/2 - x = c
or
u = c/x + x/2 + 1
and therefore
y = sqrt( c/x + x/2 + 1 ) or -sqrt( c/x + x/2 + 1 )

[@SyberMath]

Nice. Thank you!

[@leif1075]

Or why not set dy/dx equal to u and solve this way. What if you are not familiar with Bernoulli differential equation ?

[@michaelbaum6796]

Nice ode👍

[@SyberMath]

Thanks 👍

[@broytingaravsol]

very easy

[@SyberMath]

Thanks a lot 😊

[@tarunmnair]

What about the other solution ? The -ve of y ?

[@SyberMath]

It’s similar

[@tiagocantuario5424]

You missed the "+1" From "x-y^2+1" When you tried to check the positive solution :v

[@harrycomrade4580]

He noticed it

[@張茗茗-y9i]

x^2/2-xy^2+x=c

[@mcwulf25]

Was that scripted?

[@SyberMath]

Why?

[@mcwulf25]

@SyberMath The end when you didn't know then you did. Wondering if that was drama 🤔