Clever! I liked the substution which linearises your solution! I approached it like a typical 2nd order linear ODE but instead of e^mx I used x^n as its an equidimensional ODE to get a goofy auxiliary equation with complex roots for n (namely 2+/- i). Had to think for a while but x^i = e^ilnx which was a eureka moment (think complex roods of a linear 2nd order ODE)! Hence the CF is (by linearity) y=Cx^(2+i) + Dx^(2-i) = x^2(Asin(lnx)+Bcos(lnx)) Checked this solution does satisfy the homogeneous case by hand (just in case lol). And as if by magic everything cancelled! Noticed that when differentiating this function, it effectively moves the coefs around and divides by x (which is re-added by the ODE being equidimensional) which motivates the particular integral guess of the form y= x(Lsin(lnx)+Mcos(lnx)) Redefining C(x) = cos(lnx) and S(x) = sin(lnx) to save myself a headache later (noting S'(x) = C(x)/x and C'(x) = -S(x)/x). Subbing into the equation (and after a while of painful rearranging) You can find L = 1/5 and M =2/5 Hence the very elegant solution (with a clunky method) y= (x/5 + Ax^2)sin(lnx) + (2x/5 +Bx^2)cos(lnx) Thanks for the challenge had a good time battling it! (NB after checking the term equidimensional I learned today that these are called cauchy-Euler equations)
Del would actually make more sense, but he’s using Greek delta; whereas ‘del’ usually refers to (I think Hebrew) nabla, written as an inverted triangle rather than the standard triangle for Greek delta. Delta usually denotes Cauchy ‘Laplace’ operator, which is the sum of all the 2nd derivatives wrt each partial derivative.
The general formula for the multiplication of x to the k times the k-th derivative of y must be equal to Delta times (Delta - 1) times... Times (Delta - k + 1) times y
@@maths_505 Another great video. Yeah besides forgetting to go back to X which is kinda crucial tbh, it's also a good habit to check your answer by solving the original problem to test that you didn't make any mistakes. Great video!
The substitution rules out negative values of x, isn't this a problem? Edit: I didn't notice the RHS has a log(x), so x>0 was already required. But what about other cases?
Ah... brings me back... Man Cauchy and euler teaming up for a differential equation, reminds me of cauchy and Riemann collabing for differential equations...
Clever! I liked the substution which linearises your solution!
I approached it like a typical 2nd order linear ODE but instead of e^mx I used x^n as its an equidimensional ODE to get a goofy auxiliary equation with complex roots for n (namely 2+/- i). Had to think for a while but x^i = e^ilnx which was a eureka moment (think complex roods of a linear 2nd order ODE)! Hence the CF is (by linearity)
y=Cx^(2+i) + Dx^(2-i) = x^2(Asin(lnx)+Bcos(lnx))
Checked this solution does satisfy the homogeneous case by hand (just in case lol). And as if by magic everything cancelled! Noticed that when differentiating this function, it effectively moves the coefs around and divides by x (which is re-added by the ODE being equidimensional) which motivates the particular integral guess of the form
y= x(Lsin(lnx)+Mcos(lnx))
Redefining C(x) = cos(lnx) and S(x) = sin(lnx) to save myself a headache later (noting S'(x) = C(x)/x and C'(x) = -S(x)/x).
Subbing into the equation (and after a while of painful rearranging) You can find L = 1/5 and M =2/5
Hence the very elegant solution (with a clunky method)
y= (x/5 + Ax^2)sin(lnx) + (2x/5 +Bx^2)cos(lnx)
Thanks for the challenge had a good time battling it!
(NB after checking the term equidimensional I learned today that these are called cauchy-Euler equations)
Wow that's cool using D and del as notations to work with these kind of equations
Del would actually make more sense, but he’s using Greek delta; whereas ‘del’ usually refers to (I think Hebrew) nabla, written as an inverted triangle rather than the standard triangle for Greek delta. Delta usually denotes Cauchy ‘Laplace’ operator, which is the sum of all the 2nd derivatives wrt each partial derivative.
@@brabhamfreaman166 ok bro 👍
One similar thing I like is when people use prime for d/dx and the dot for d/dt
Wow, very cool operator manipulation. I‘ve never seen this before. I have learned a lot. Thank you👍
11:30
Should be in terms of x
y=x²(Acoslnx+Bsinlnx)
+x(2/5 coslnx +1/5sinlnx)
A much substutution would have been e^(-3t).
The general formula for the multiplication of x to the k times the k-th derivative of y must be equal to Delta times (Delta - 1) times... Times (Delta - k + 1) times y
Delta is used in so called finite calculus
as difference operator
Indeed
You didn't finish. need go back to x
Shouldn't finish too early 😂
Jokes aside yeah I forgot
@@maths_505 Another great video. Yeah besides forgetting to go back to X which is kinda crucial tbh, it's also a good habit to check your answer by solving the original problem to test that you didn't make any mistakes.
Great video!
@@zunaidparker thanks mate
Do you have a video explaining why operational calculus works? Something like cancelling the y at 2:46 seems wrong to me, but it works out doesn’t it?
The substitution rules out negative values of x, isn't this a problem?
Edit: I didn't notice the RHS has a log(x), so x>0 was already required. But what about other cases?
Nice
I came here to say that u spelled Euler wrong :)
What's the difference between me and a calandar
A calculator can't watch this awesome DE video😂
No
A calandar has dates
@@hamdanrox 🤣🤣🤣
😭🤣🤣🤣
Ah... brings me back... Man Cauchy and euler teaming up for a differential equation, reminds me of cauchy and Riemann collabing for differential equations...
I'd just go full frobenious on these bad boys
SUIIIIIIIIIIIIIIII