too tricky for your differential equations class??
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- เผยแพร่เมื่อ 20 ก.ย. 2024
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That's just a Bernoulli differential equation, I did learn about it in my differential equations course
Yes , it is easy to make it Ricatti and then it will be a little bit challenging
Guess particular solution and use substitution if you want to get Ricatti equation from Bernoulli
then you can give your fellow to solve it
Y=x is a solution too
which is C = 0, yes
His general solution includes that
Actually, there were two errors/omissions in this video that canceled each other out. First, the solutions y=x (z=1) and y=z=0 were missed in the first equation of the right side of the board when both sides were divided by z(z-1) without checking first when the denominator equals 0. However, in the transition from the second-to-last to the last equation, the solution y=x magically reappeared after substituting exp(c) with C (let's call it C_1). This is because there is no c such that exp(c)=0, but C_1 can be 0. Still, y=0 solution is still missing.
@@dadjokes2815 actually nothing is missed, rearranging gives solution:
y = x/(1-Cxe^(x^2/2))
Taking C = 0 gives the solution y = x and letting C -> ±∞ gives the trivial solution
y == 0
with the substitution z=1/y, the o.d.e becomes standard.
5:54
is where the video starts
Upon 5s inspection, it is a Bernoulli/Ricatti type. 10s inspection suggests z=xy as a reducing transform to separable ODE
Doing the algebra to solve for z and y you get the following general solutions. (I think the sign of C has been switched compared to the video.)
z = 1 / [1 + Cxe^(1/2*x^2)]
y = x / [1 + Cxe^(1/2*x^2)]
At the end, I think it would be easier to turn the (y-x) / y inside the log into 1 - x/y right away and then just exponentiate both sides, take the reciprocal of both sides, and multiple both sides by -x.
Homework, great!😂
My method diverged from yours at the point when you made the second substitution.
I used partial fractions to integrate the left side (the 1/z(z-1) part). Also I didn't translate z back into y until the very end.
I ended up with y = - x / (x * e^(1/2 x + c) +1)
In my diff eq course we learned equations of this type, where the substitution u=y/x makes them separable. They were called homogeneous differential equations
Divide by y² and solve
I feel the replacement of z by y/x at 4:40 was too early - solving for z first seems easier?
Potentially if you want to rearrange directly for y it’s quicker, I suppose it doesn’t make that big of a difference in the end
Easy , this is Bernoulli type equation
We can use variation of parameter directly to the Bernoulli equation
If we prefer integrating factor there also is quite easy to find
and it is in the form mu(x,y) = phi(x)psi(y)
y'-y/x=y²/x²+y²-xy-y/x (Doing this to the left side forcibly to bring something reasonable)
y'-y/x=y²(1+1/x²)-xy(1+1/x²)=(1+1/x²)(y²-xy)=y(y-x)(1+1/x²)
Now y'/x-y/x²=d(y/x)/dx=(y/x)(y-x)(1+1/x²)=(y/x)(y/x-1) (x)(1+1/x²)
now d(y/x)/(y/x)(y/x-1)=x(1+1/x²)dx
tbh i don't like ODEs solved in this fashion, these problems are way degenerate compared to the whole set of all ODEs - where we have to find numerical solutions most of the time if known standard techniques don't work
/y^2... y'/y^2=1/x^2+1+x/y... t=1/y... t'= -y'/y^2... -t'=1/x^2+1-xt... u=xt...t=u/x...t'=(u'x-u)/x^2... u-u'x=1+x^2-ux^2... /x... u/x-u'=1/x+x-ux... -u'=(1/x+x)-u(1/x+x)... u'/(u-1)=1/x+x... ln|u-1|=lnx+x^2/2+C...u=exp(lnx+x^2/2+C)+1...u=x*exp(x^2/2+C)+1...t=exp(x^2/2+C)+1/x... y=x/((x*exp(x^2/2+C)+1)
this is correct, I confirm.
@@Andrew-II thank you!
fourth to post! I went forth to post and became fourth to post= Yay!
Anyway ~ collywobbles? Original equation as presented gives y' = 0 when y = 0 AND x0 as it is a divisor in posed equation. EDIT: y=0 holds for all x
But that sort of contradicts presence of y as a divisor in solution presented at 5:41
This is not to say boo or boo hoo. It more or less says Yay! but what happened and why a contradictory event?
ps: placing a y0 at 0:00 is not allowed