Calculus | VLR.gg

#1

ArtemisZ083

I need help with the integration of this dy/dx = x - 2y/x
i tried to take help from ai but this is what ir proved 💀
The correct answer is (3) (-1,2).

Given the slope of the tangent at any point (x, y) on the curve as (x²-2y)÷x, we can write the differential equation:

dy/dx = (x²-2y)/x

We are also given that the curve passes through the point (1, -2). We can use this information to find the equation of the curve.

Separating the variables, we get:

∫dy = ∫(x²-2y)/x dx

Integrating both sides, we get:

y = (1/3)x³ - 2x + C

Using the point (1, -2), we can find the value of C:

-2 = (1/3)(1)³ - 2(1) + C
C = -4/3

So, the equation of the curve is:

y = (1/3)x³ - 2x - 4/3

Now, we can check if the curve passes through the point (-1, 2):

2 = (1/3)(-1)³ - 2(-1) - 4/3
2 = -1/3 + 2 - 4/3
2 = 2/3 (which is true)

Therefore, the curve also passes through the point (-1, 2).

#2

ButterflyEffect23

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ok thanks

#3

ArtemisZ083

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Do you know calculus?

#4

Dreoxx

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what do you need help with

#5

ArtemisZ083

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dy/dx = x - 2y/x integration of this

#8

Dreoxx

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dy/dx + P(x)y = Q(x)

dy/dx + 2y/x = x

P(x) = 2/x and Q(x) = x

btw integrating factor is e^(int of P(x) dx)

I expect you to solve it from here

#10

Dreoxx

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the second line (dy/dx + 2y/x = x) is a rearranged format of dy/dx = x - 2y/x to match the first line (dy/dx + P(x)y = Q(x))

#12

ArtemisZ083

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i have no idea what any of this means, can you please explain? We havent been taught integration yet so idk what an integrating factor is, i only know basic integration rules that are used in physics for taking elements

#16

Flos

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Search up first order differential equations on yt

#18

Dreoxx

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why is your instructor trying to make you solve something you haven't learned?

so this is a first order linear diff. equation.

dy/dx + P(x)y = Q(x) is called the standard form in this case, so we will try to reformat the initial equation so we can get P(x) and Q(x), which will be used later on in the process.

#19

ArtemisZ083

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I was actually just solving a previous year questions of this chapter, teacher didnt give us the question.

#22

Dreoxx

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the integration factor i mentioned earlier, e^(int of P(x) dx), is then multiplied on both sides of the differential equation we reformatted to match the standard form.

in this case it would be x^2 times dy/dx + x^2 times 2y/x = x^2 times x (we have multiplied x^2 to every term), which simplifies into x^2 dy/dx + 2xy = x^3

from there we find what the left side would be the derivative of:

d/dx of yx^2 would be the left side, since it would become x^2 + 2xy (use the concepts of chain, product rule)

so it becomes d(yx^2)/dx = x^3, and integrating both sides becomes

yx^2 = x^4/4 + C, simplifying into y = x^2/4 + C/x^2

plug in initial values, and there you go

#23

Dreoxx

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personally I got (1, -2) as well so your answer key is bugging

#25

ArtemisZ083

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Actually the question says (1,-2) does satisfy from that we find the constant, and then we have to find another point that satisfies, from the given options

#24

ArtemisZ083

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Thanks a lot man, ITS CORRECT <333

#26

Dreoxx

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no problem

#27

ArtemisZ083

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this is the whole ques
if a curve passes through the point (1,-2) and has slope of the tangent at any point (x,y) on it as (x²-2y)÷x, then the curve also passes through the point: (1) (3,0) (2) (√3,0) (3) (-1,2) (4) (-√2,1)
Now slope means differentiation and if we integrate the slope we get the curve and to find a constant we make use of the point given. But how do i integrate this

#28

Dreoxx

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um did you accidentally copy paste the question again while replying to me

continue thread →

#40

archetype

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when i came into this thread asking for calculus i did not expect it to be a diffeq question lol i expected like calc 1 or 2 or 3

#41

Dreoxx

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diff eq is part of calc 2 iirc

#45

archetype

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ah, i meant like ODE, which is a different course at my institution. The course that teaches Wronskian

#46

Dreoxx

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im not too familiar with these terms (ODE, Wronskian) since I've undergone my education in America

#47

archetype

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Yeah I'm also an overseas student in the US lmao
ODE = Ordinary Differential Equations

You can google Wronskian Equations and there's an easy image that'll explain what they are.

#50

Dreoxx

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i see, im not too into diff eq, I myself am an aspiring psych major and I'm still in high school

#51

archetype

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yeah you're prob never gonna learn it then lmfao it's a college course and afaik is only for engineering/stem students

continue thread →

#6

NexusNomad

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Take y/x = v and solve it. This is one of the most basic questions in differential equations.

#7

ArtemisZ083

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yes we havent been taught it yet, we are on AOD rn can you give the full solution

#9

NexusNomad

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y/x = v
y=xv
dy/dx = v + x*dv/dx

Substitute in the given equation we get

v + xdv/dx= x - 2v

Solve it from here on your own I'm too lazy and replace v= y/x in the end.

#17

ArtemisZ083

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this is the whole ques
if a curve passes through the point (1,-2) and has slope of the tangent at any point (x,y) on it as (x²-2y)÷x, then the curve also passes through the point: (1) (3,0) (2) (√3,0) (3) (-1,2) (4) (-√2,1)
Now slope means differentiation and if we integrate the slope we get the curve and to find a constant we make use of the point given. But how do i integrate this

#20

NexusNomad

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Ok I'll do one more step

dv= dx - 3v dx/x

Integrate it on both sides we get

v= x + 3v/x^2 + C

Substituting v = y/x we get

y/x = x + 3y/x^3 + C

Given it passes through (1,-2) we Substitute this in the above equation to get the value of C

Do it from here or should I do that for you too?

#21

ArtemisZ083

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No thanks a lot nexus, i appereciate it a lot <3 Btw do you know you have the most disliked post on the internet?

Your solution is wrong

#34

NexusNomad

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Yeah mb, you should find the integration factor and do it like the guy above. I don't even use this anymore so I just randomly used the other method. Anyways if you have doubts about jee you can ask me, trust me I'm in one of those big colleges.

#36

ArtemisZ083

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Its okay man thanks for the help, in which iit are you?

#11

DSGFan

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Your post is a bit unclear.

You have the equation dy/dx = x - 2y/x.

Is the goal to find y(x)? Were you given any initial conditions?

#13

Dreoxx

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I'm assuming he was given initial conditions and he's tryna find the specific solution under the general +C format, bc he's verifying whether it passes a point

#15

ArtemisZ083

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this is the whole ques
if a curve passes through the point (1,-2) and has slope of the tangent at any point (x,y) on it as (x²-2y)÷x, then the curve also passes through the point: (1) (3,0) (2) (√3,0) (3) (-1,2) (4) (-√2,1)
Now slope means differentiation and if we integrate the slope we get the curve and to find a constant we make use of the point given. But how do i integrate this

#14

ArtemisZ083

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yes this is the whole ques
if a curve passes through the point (1,-2) and has slope of the tangent at any point (x,y) on it as (x²-2y)÷x, then the curve also passes through the point: (1) (3,0) (2) (√3,0) (3) (-1,2) (4) (-√2,1)
Now slope means differentiation and if we integrate the slope we get the curve and to find a constant we make use of the point given. But how do i integrate this

#30

Dreoxx

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i see yeah i got the question now

my explanation should still stand just plug in choices (brute force since its multiple choice) and you get the correct one

#31

h1gh_fall

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how can i

#38

ArtemisZ083

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Its okay i already have the solution now, thanks for trying <3

#60

h1gh_fall

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i didn't try i was going to write something but i forgor

#32

adi02

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Fuck, this feels old, been 5 years

#37

ArtemisZ083

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No problem lol

#33

adi02

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You preparing for JEE?

#35

ArtemisZ083

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I am

#39

adi02

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Good luck! I am an IITian, always here if u need any help

#44

ArtemisZ083

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Thank you man sure

#42

rStellar

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give up on JEE ima beat u anyway

#43

Dreoxx

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with that attitude even if you become AIR 1 you will achieve nothing in life

#48

i11matic

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Imagine thinking math is real

#49

Froggy0_0

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imagine thinking

#52

butterdog_dogwithdabutter

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1-2y/x leke kar
1 constant me chala jayega aur fir u/v laga de