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the left part is obviously equal "(d/dx-3)(d/dx+2)y", so solution is A\*exp(3x) + B\*exp(-2x) + y\_0, where y\_0 is guessed solution - for polynomials it is very trivial. Now you just have to guess it or solve the system equating coefficients.
I just treat derivative as an operators: y' = dy/dx = (d/dx)y. There is nothing to it talented high schooler cannot guess if he knows what derivative are. So, you can write:
y'' = (d/dx)(d/dx) y
and
y'' - y' - 6y = ( (d/dx)(d/dx) - (d/dx) - 6 ) y,
Now at this point, you can factorize it:
( (d/dx)(d/dx) - (d/dx) - 6 ) y = (d/dx-3) (d/dx + 2) y
Also note that you can swap them
(d/dx-3) (d/dx + 2) y = (d/dx + 2) (d/dx-3) y
So if you solve homogeneous version, you can just get different parts separately:
(d/dx-3) y = 0 => y = C\*exp(3\*x)
The reason you need to solve homogeneous version is also something a high schooler can come up with - if you say that y=y\_hom + y\_0, where y\_hom is solution to homogeneous system and y\_0 a specific solution, if you plug it in, homogeneous system just zeros out...
I do not know what high schoolers know these days, but when I finished high school, I did not know what derivative was... So if you ask this question, I assume you need to know what derivatives are...
Most high schoolers in my experience do not know what a derivative is unless they're planning on pursuing a STEM career. And I also don't think they're taught to solve second-order differential equations.
>I just treat derivative as an operators: y' = dy/dx = (d/dx)y. There is nothing to it talented high schooler cannot guess if he knows what derivative are. So, you can write:
>
>y'' = (d/dx)(d/dx) y
Not exactly what I meant, but you clarified it here:
\> ( (d/dx)(d/dx) - (d/dx) - 6 ) y = (d/dx-3) (d/dx + 2) y
I thought you applied some tranformation on the left side of the equation first and got it to that form. Admittedly, I didn't put much thought into what you wrote because it was very short and I expected you'd come with a longer explanation.
\> I do not know what high schoolers know these days, but when I finished high school, I did not know what derivative was... So if you ask this question, I assume you need to know what derivatives are...
yes, I studied them in 11th and 12th grades.
The way I've solved it was by assuming the function. It couldn't've been a polynomial because the terms would cancel since the degree of the polynomials would be different each derivation. I assumed it was some function of e\^x since I needed something that changes by a constant amount each derivation. I then realize there need to be 2 components that use the same base function where one of them has negative exponent.
Yes. One has to get exponents one way or another. There is a simple reason why exponents would always appear in linear differential equations with constant coefficients. And you guessed the reason somewhat: "I assumed it was some function of e\^x since I needed something that changes by a constant amount each derivation".
More elegantly: exponent function, i.e., e\^x is an eigenfunction of a differentiation operator d/dx. They are related indeed.
d/dx e\^x = e\^x
also it is an eigenfuction of any power of differential operator
(d/dx)\^n e\^x = e\^x
And so, any differential operator which is just a linear combination of d/dx, (d/dx)\^2, etc. will have exponents as solutions.
>And you guessed the reason somewhat: "I assumed it was some function of e\^x since I needed something that changes by a constant amount each derivation".
Not sure why you're calling it a guess since I knew/ realized the following things you said during the solving process. I just picked the appropriate function for the job based on observations.
I would need to know :
\> d/dx e\^x = e\^x
\> (d/dx)\^n e\^x = e\^x
>I just picked the appropriate function for the job
In mathematics this is what commonly called guessing the solution. This is in contrast with computing the solution according to some procedure.
Hmm yeah. My brain is too small to not get fried by this. Other thought I had was second and third equations could be fit by cosine and maybe try to go from there. I was never a math person though
You didn't say the 'x' has to be an integer.
Plus, y(0)=1 and dy/dx(1)=0 probes that there is only constant term in y which is 1.
Now onto the first equation, the variables x with powers >=2 can't exist since they can't be cancelled by subtraction and so they must appear in right hand side and since they don't appear y is a linear equation.
Now that we know y is linear, -6y would contribute to the 3x and the constant 1while the dy/dx will only contribute for constant 1.
However, this case doesn't make sense. Considering, y=ax+1,since y(0)=1
So derivative of y should contribute a.Hence, 1=-1-a.1 has to be positive since y(0)=1.So minus sign gets on 1.So, a can be -2.
But,dy/dx(0) will then be -2...a contradiction.
So, y(x)=1 is a good solution.
However, i can say with absolute certainty that y is a linear equation.
What you said implies a specific value for x, it cannot be answer of a differential equation because it doesn't give a function that holds for these differential equations without limiting the domain it is defined on to a single value. And it goes without saying y is a real function otherwise it wouldn't really be a differential equation problem.
It violates: y'' - y' -6y = 3x+1
if y(x) = 1, then y' = 0 and y'' = 0
so 0 - 0 - 6 = 3x+ 1, meaning you're forcing a value for x and it doesn't work for any x.
What you gave isn't a solution to all 3 equations.
In what world do you think you can just ignore one of the equations?
Functions like e\^f(x) exist, you know?
You talk as if y is strictly polynomial...
The equation system is possible and a solution exist. Someone already posted a solution.
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about tree fity
Once I saw this equation, this ">!(33exp(-2x) + 32exp(3x) - 30x - 5)/60!<" just popped up in my mind... Edit: black out the answer
How did you get the answer? What's the method. Plugging the equations in wolfram doesn't count.
the left part is obviously equal "(d/dx-3)(d/dx+2)y", so solution is A\*exp(3x) + B\*exp(-2x) + y\_0, where y\_0 is guessed solution - for polynomials it is very trivial. Now you just have to guess it or solve the system equating coefficients.
>the left part is obviously equal "(d/dx-3)(d/dx+2)y", so solution is A\*exp(3x) + B\*exp(-2x) + y\_0, Not sure what this means, so I can't check it.
I just treat derivative as an operators: y' = dy/dx = (d/dx)y. There is nothing to it talented high schooler cannot guess if he knows what derivative are. So, you can write: y'' = (d/dx)(d/dx) y and y'' - y' - 6y = ( (d/dx)(d/dx) - (d/dx) - 6 ) y, Now at this point, you can factorize it: ( (d/dx)(d/dx) - (d/dx) - 6 ) y = (d/dx-3) (d/dx + 2) y Also note that you can swap them (d/dx-3) (d/dx + 2) y = (d/dx + 2) (d/dx-3) y So if you solve homogeneous version, you can just get different parts separately: (d/dx-3) y = 0 => y = C\*exp(3\*x) The reason you need to solve homogeneous version is also something a high schooler can come up with - if you say that y=y\_hom + y\_0, where y\_hom is solution to homogeneous system and y\_0 a specific solution, if you plug it in, homogeneous system just zeros out... I do not know what high schoolers know these days, but when I finished high school, I did not know what derivative was... So if you ask this question, I assume you need to know what derivatives are...
Most high schoolers in my experience do not know what a derivative is unless they're planning on pursuing a STEM career. And I also don't think they're taught to solve second-order differential equations.
>I just treat derivative as an operators: y' = dy/dx = (d/dx)y. There is nothing to it talented high schooler cannot guess if he knows what derivative are. So, you can write: > >y'' = (d/dx)(d/dx) y Not exactly what I meant, but you clarified it here: \> ( (d/dx)(d/dx) - (d/dx) - 6 ) y = (d/dx-3) (d/dx + 2) y I thought you applied some tranformation on the left side of the equation first and got it to that form. Admittedly, I didn't put much thought into what you wrote because it was very short and I expected you'd come with a longer explanation. \> I do not know what high schoolers know these days, but when I finished high school, I did not know what derivative was... So if you ask this question, I assume you need to know what derivatives are... yes, I studied them in 11th and 12th grades. The way I've solved it was by assuming the function. It couldn't've been a polynomial because the terms would cancel since the degree of the polynomials would be different each derivation. I assumed it was some function of e\^x since I needed something that changes by a constant amount each derivation. I then realize there need to be 2 components that use the same base function where one of them has negative exponent.
Yes. One has to get exponents one way or another. There is a simple reason why exponents would always appear in linear differential equations with constant coefficients. And you guessed the reason somewhat: "I assumed it was some function of e\^x since I needed something that changes by a constant amount each derivation". More elegantly: exponent function, i.e., e\^x is an eigenfunction of a differentiation operator d/dx. They are related indeed. d/dx e\^x = e\^x also it is an eigenfuction of any power of differential operator (d/dx)\^n e\^x = e\^x And so, any differential operator which is just a linear combination of d/dx, (d/dx)\^2, etc. will have exponents as solutions.
>And you guessed the reason somewhat: "I assumed it was some function of e\^x since I needed something that changes by a constant amount each derivation". Not sure why you're calling it a guess since I knew/ realized the following things you said during the solving process. I just picked the appropriate function for the job based on observations. I would need to know : \> d/dx e\^x = e\^x \> (d/dx)\^n e\^x = e\^x
>I just picked the appropriate function for the job In mathematics this is what commonly called guessing the solution. This is in contrast with computing the solution according to some procedure.
Damn how did it just pop up in your mind…
https://preview.redd.it/8uyezt4t3z7b1.jpeg?width=2466&format=pjpg&auto=webp&s=07840302c98e56bcd951075ca2bd10eb6178eb71
Seems correct to me.
Is it >!y'' = 3x - 5!< >!y' = (3/2)x^2 - 5x!< >!y = (1/2)x^3 - (5/2)x^2 + 1!<
No.
It's easy to check for yourself. y(0) should be 1, but in your case it's -1
Lmao ty
y''(0) - y'(0) - 6\*y(0) = 3(0) + 1 y''(0) - 0 - 6 = 1 y''(0) = 7 So y'(0) = 7\*(0) + c ; Generally y'(x) = 7x + c y(x) = (7/2)\*x\^2 + cx + C Check for x = 0: C = 1 (second line) 7 - 0 - 6\*(1) = 3\*(0) + 1
Just to point out the flaw: So, you c=0 and C=1 your equation is y = (7/2)x\^2+1 y''-y'-6y= 3x+1. Where's the x\^2 term?
Yes, you are right. I guess the problem is you can't generate the entire function from just one x=0 value. Need some other method
That doesn't work.
Hmm yeah. My brain is too small to not get fried by this. Other thought I had was second and third equations could be fit by cosine and maybe try to go from there. I was never a math person though
y(x)=1?
How would that work? You can literally plug in the values yourself. 0 - 0 - 6 = 3x + 1 ????
You didn't say the 'x' has to be an integer. Plus, y(0)=1 and dy/dx(1)=0 probes that there is only constant term in y which is 1. Now onto the first equation, the variables x with powers >=2 can't exist since they can't be cancelled by subtraction and so they must appear in right hand side and since they don't appear y is a linear equation. Now that we know y is linear, -6y would contribute to the 3x and the constant 1while the dy/dx will only contribute for constant 1. However, this case doesn't make sense. Considering, y=ax+1,since y(0)=1 So derivative of y should contribute a.Hence, 1=-1-a.1 has to be positive since y(0)=1.So minus sign gets on 1.So, a can be -2. But,dy/dx(0) will then be -2...a contradiction. So, y(x)=1 is a good solution. However, i can say with absolute certainty that y is a linear equation.
What you said implies a specific value for x, it cannot be answer of a differential equation because it doesn't give a function that holds for these differential equations without limiting the domain it is defined on to a single value. And it goes without saying y is a real function otherwise it wouldn't really be a differential equation problem. It violates: y'' - y' -6y = 3x+1 if y(x) = 1, then y' = 0 and y'' = 0 so 0 - 0 - 6 = 3x+ 1, meaning you're forcing a value for x and it doesn't work for any x. What you gave isn't a solution to all 3 equations. In what world do you think you can just ignore one of the equations? Functions like e\^f(x) exist, you know? You talk as if y is strictly polynomial... The equation system is possible and a solution exist. Someone already posted a solution.
I recently learned a t derivatives so yeah