Solving A Non-Linear Differential Equation
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if you do u=y^2+x you end up with a separable equation
Cool!
Please more differential equations đ
When that alarm went off thought you'd added cool outro music to your video đ
u' - 2u = 2x looks like a situation to use integrating factor, with I = e^-2x - wonder if we arrive at the same conclusion. granted we would have to do integration by parts on the RHS.
Your method was more complicated than needed.
This is a bernuli equation, if you put it in the form
y' - y = x*(y^-1)
Bernuli happens when you have a y^1 term and y^n term
You do the sub u=y^(1-n), in this case y^(1- (-1)) or u=y^2
From this du/dx = 2y dy/dx
Now just multiply the original equation by our u and you get
2y* dy/dx '- 2y^2 =2x
Notice the diff we did earlier so sub in du/dx = 2y dy/dx
We now have du/dx - 2u =2x
This is a linear in u, the integration factor ends up being e^(-2x) and when we clean up we get u*e^(-2x) = 2* the integral [ x*e^(-2x) ] dx
Do int by parts then substitute back in u=y^2 and the final result is
y = ± sqrt[A*e^(2x) - x - œ]
At the first sight I see Bernoulli equation
Bernoulli equation has its own integrating factor
mu(x,y) = exp((1-r)Int(p(x),x))y^{-r}
so it is not necessary to reduce it to the linear equation
Procedure for solving Bernoulli equation is similar to this which is used for solving linear equation
Too complicated! đ€Șđ
A cool substitution đ
Glad to hear that
@@SyberMathtry this problem
Find a function where the derivative is equal to its inverse fâ(x)=f^-1(x)
I thought you were about to do a W at the end there.
@4:50 I think the usage of D is right but the common part is not, I think it should be (D-2)u=0, thus, we can replace D with r to get homogeneous equivalent equation... but somehow the solution arrived was the same... đ đ đ
Ooopsies! đ
do you know ( or anyone that read this comment) know any programs I can have graphs of x,y and Im at the same time? on the internet its called phantom graphs
what is the height of a fluid inside a sphere when it occupies 1/4 of the sphere?
â
Calculate it.
Are we ever gonna run out of problems to solve?
Probably not