A Non-Homogeneous Differential Equation
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I solved it a different way. Bringing y over to the LHS: y’ - y = x^2
I noticed that the derivative of ye^-x is (y’-y)e^-x. So, if we multiply both sides by e^-x, we get:
d(ye^-x)/dx = x^2 e^-x
Then we just need to integrate the RHS (e.g., using integration by parts).
Very good!
You have such a nice voice it makes your explanations pleasant to the ears
Thank you for the kind words!!! 🧡🧡
I just used the method of variation of an arbitrary constant.
Can be generalized to y'= y + p(x), where p(x) is a polynomial function.
The more interesting problem you obtain after replacing y with x but only on the RHS.
So funny, that the 2 different constants, both called c, do not interfere with each other in the story (by accident).
I have only studied limits and derivatives till now but still it was easy to understand this video
Thank you for making maths as entertaining as it is supposed to be
Use the general form, dy/dx +P(x)y = Q(x) to find u(x) = e^-x the integrating factor. Use u(x) to determine the general solution by multiplying both sides of equal sign by u(x), left side of equal sign is the derivative of the product of u(x) and y(x), integrate both sides of equal sign and simplify to find y(x).
y(x)=a e^x + b x² + c x +d.
put y in the original equation and find a, b, c, d. Finished.
Nice
Y = -x^2-2x-2+ce^x, linear first order deq, nice job
Yes! Thank you!
You could also say y=c*e^x - (x+1)^2 - 1.
Isn't this just Bernoullis lijear differential equation , find the integrating factor that is e^-x and then find solution by y.e^-x = integral of x².e^-x and solve using by parts
Well, this is a linear differential equation, so we can solve it that way too
Can also be solved using the laplace transform
A curiosity: I took my polynomial exponents in the other direction and found f(x) = c*e^x + x^3/3 + x^4/12 + x^5/60 + ... But that is just 2*e^x minus the first 3 terms. So then I got: y = c*e^x + 2e^x - (x^2 + 2x + 2)
or if k = c+2, y = ke^x - (x^2 + 2x + 2). So I backed into the correct answer having started the polynomial towards an infinite series.
Sorry, can you help me with that exercice? Thank you! Three numbers, the sum of which is 114, can be considered as three consecutive limits of a geometric progression, or as the first, fourth and twenty-fifth limits of an arithmetical progression. Find these numbers.
by limits I assume you mean the first, fourth and twenty-fifth terms of the progression, right?
@@themathsgeek8528 naturally
@@klementhajrullaj1222et the numbers be A,B and C (in that order in the AP and GP)
Since they are consecutive terms of a GP, we can write the ratio between them as C/B=B/A which gives AC=B^2
Now since they are the first, fourth and twenty-fifth terms of an AP, the common difference of that AP can be written as
(B-A)/3 = (C-B)/21
Which upon simplification gives 8B = 7A+C...
Now we can construct a quadratic with roots 7A and C since we know their product (7B^2) and sum(8B)..
So 7A and C are roots of x^2-8Bx+7B^2=0
This upon solving gives 7B and B as the roots..
So now one case would be where 7A=7B and B=C which gives A=B=C=38 since their sum is 114.
The other case gives 7A=B and C=7B=49A
Using the fact that the numbers sum to 114, we can easily obtain them to be 2,14,98 in that order.
Hence the numbers can be all 38 or (2,14,98).
Hope this is clear.
The ap would be one with 4n-2 as its nth term
Let*
brought me back to 2004, university 2nd year. thanks SyberMath 🤗
Did you go on to work in a stem field? If so, which one? Thanks.
ce^x-(x²+2x+2)
Solution:
dy/dx = y+x²
------------
Substitution: z=y+x² z’=y’+2x y’=z’-2x
-------------
z’-2x = z |+2x ⟹
z’ = z+2x
------------
Substitution: u=z+2x u’=z’+2 z’=u’-2
-------------
u’-2 = u |+2 ⟹
du/dx = u+2 |*dx/(u+2) ⟹
du/(u+2) = dx |∫() ⟹
ln|u+2| = x+C |e^() ⟹
u+2 = e^(x+C) = K*e^x |Resubstitution: u=z+2x ⟹
z+2x+2 = K*e^x |Resubstitution: z=y+x² ⟹
y+x²+2x+2 = K*e^x |-x²-2x-2 ⟹
y = K*e^x-x²-2x-2
Sample:
left side: y’ = K*e^x-2x-2
right side: y+x² = K*e^x-x²-2x-2+x² = K*e^x-2x-2 ⟹
both sides are equal, all o.k.
i love all your videos
Thank you! 🥰
Ok, this is a tough one to just do in my head, but I think the solution is… [SPOILER]
y = -x^2 - 2x - 2 + Ce^x
(whew!)
Nice!!!
Haaa
Deep math begin to appear
Unfortunate choice to use "c" for the constant in the homogeneous solution and also as the constant term for the particular solution. These should be independent values.
y=-x^2-2x-2+ce^x...si può calcolare a mente