/sci/ - Science & Math » Thread #10862404

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Anonymous Sun 04 Aug 07:00:06 2019 No.10862404 View Reply Original Report

Quoted By: >>10862407 >>10862479 >>10862481 >>10862772 >>10863472 >>10863498 >>10864795

Can you solve a system of differential equations with substituting one of the equations back in? like once you get the equation for x, you can plug that in for x into one of the equations and solve for y. but can you not also solve for y just by elimination the same you did for x? for some reason they don't give the same answer.

Anonymous

Anonymous Sun 04 Aug 2019 07:00:52 No.10862407 Report

Quoted By:

>>10862404
I mean WITHOUT substituting one of the equations back in*

Anonymous

Anonymous Sun 04 Aug 2019 07:19:31 No.10862431 Report

Quoted By:

what im basically saying is, once you get a solution for one variable, lets say x - you can get the answer for y by plugging in for x. but do you HAVE to solve it this way when doing elimination or can you just solve for y like you did for x

Anonymous

Anonymous Sun 04 Aug 2019 07:49:06 No.10862473 Report

Quoted By:

bump

Anonymous

Anonymous Sun 04 Aug 2019 07:53:23 No.10862479 Report

Quoted By: >>10862519

>>10862404
just because you get a different answer doesn't mean it's wrong
you can always check your answers by just plugging them into the original equations and seeing if they work

Anonymous

Anonymous Sun 04 Aug 2019 07:54:26 No.10862481 Report

Quoted By:

>>10862404
>for some reason they don't give the same answer.

That means that you fucked up.

Anonymous

Anonymous Sun 04 Aug 2019 08:30:56 No.10862519 Report

Quoted By: >>10862547

>>10862479
so it is okay to solve them both by elimination?

Anonymous

Anonymous Sun 04 Aug 2019 09:01:41 No.10862547 Report

Quoted By: >>10862920

>>10862519

the value of x and y doesn't change, it doesn't matter how you solve the problem

Anonymous

Anonymous Sun 04 Aug 2019 10:52:45 No.10862772 Report

Quoted By:

>>10862404
why should your system even have a solution?

Anonymous

Anonymous Sun 04 Aug 2019 12:58:24 No.10862920 Report

Quoted By: >>10863104

>>10862547
>the value of x and y doesn't change, it doesn't matter how you solve the problem
what did he mean by this

Anonymous

Anonymous Sun 04 Aug 2019 15:19:37 No.10863104 Report

Quoted By:

>>10862920
clarification: the general solution for any system of equations is always the same regardless of the method used to find it

Anonymous

Anonymous Sun 04 Aug 2019 19:38:12 No.10863472 Report

Quoted By:

>>10862404
>for some reason they don't give the same answer.
Are you sure about that?
I haven't tried solving the system in your image, but I recall being quite stumped initially when I tried solving $\displaystyle \int \cos{(x)}\sin{(x)}\,dx$ and got the "different" answers of $\displaystyle \frac{1}{2}\sin^2{(x)} + C$ , $\displaystyle \int -\frac{1}{2}\cos^2{(x)} + C$ , and $\displaystyle \int -\frac{1}{4}\cos{(2x)} + C$ depending on how I solved.
Of course, I eventually realized that all the answers were the same. $\displaystyle \frac{1}{2}\sin^2{(x)} = \frac{1}{2} - \frac{1}{2}\cos^2{(x)} = \frac{1}{4} - \frac{1}{4} \cos{(2x)}$ . From that day onward, I kept with me an important lesson: never overlook the constants.
Have you overlooked the constants, my indifferent friend?

Anonymous

Anonymous Sun 04 Aug 2019 19:50:36 No.10863498 Report

Quoted By:

>>10862404
>Can you solve a system of differential equations with substituting one of the equations back in?
I don't see why not.

Anonymous

Anonymous Mon 05 Aug 2019 04:04:05 No.10864795 Report

Quoted By:

>>10862404
yes

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