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.
@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... 😅😅😅
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
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 - ½]
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
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if you do u=y^2+x you end up with a separable equation
Please more differential equations 😊
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.
@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... 😅😅😅
Ooopsies! 😂
When that alarm went off thought you'd added cool outro music to your video 😊
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! 🤪😁
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 - ½]
Cool!
what is the height of a fluid inside a sphere when it occupies 1/4 of the sphere?
Calculate it.
A cool substitution 👍
Glad to hear that
@@SyberMathtry this problem
Find a function where the derivative is equal to its inverse f’(x)=f^-1(x)
d^2phi/dx^2=1/etea^2(phi^3-phi)
How solve this ode for phi
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
How do we know this is the only solution?
I thought you were about to do a W at the end there.
Are we ever gonna run out of problems to solve?
Probably not