Calculus

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).

ok thanks

ButterflyEffect23 [#2]

ok thanks

Do you know calculus?

what do you need help with

Dreoxx [#4]

what do you need help with

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

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

NexusNomad [#6]

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

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

ArtemisZ083 [#5]

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

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

ArtemisZ083 [#7]

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

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.

Dreoxx [#8]

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

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))

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?

Dreoxx [#8]

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

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

DSGFan [#11]

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?

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

DSGFan [#11]

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?

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

Dreoxx [#13]

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

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

ArtemisZ083 [#12]

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

Search up first order differential equations on yt

NexusNomad [#9]

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.

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

ArtemisZ083 [#12]

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

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.

Dreoxx [#18]

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.

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

ArtemisZ083 [#17]

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

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?

NexusNomad [#20]

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?

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

Dreoxx [#18]

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.

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

Dreoxx [#22]

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

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

Dreoxx [#22]

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

Thanks a lot man, ITS CORRECT <333

Dreoxx [#23]

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

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

ArtemisZ083 [#24]

Thanks a lot man, ITS CORRECT <333

no problem

Dreoxx [#26]

no problem

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

ArtemisZ083 [#27]

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

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

Dreoxx [#28]

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

Nope i was just giving you the question as you seemed confused about the answer key

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

how can i

Fuck, this feels old, been 5 years

You preparing for JEE?

ArtemisZ083 [#21]

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

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.

adi02 [#33]

You preparing for JEE?

I am

NexusNomad [#34]

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.

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

adi02 [#32]

Fuck, this feels old, been 5 years

No problem lol

h1gh_fall [#31]

how can i

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

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

Dreoxx [#8]

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

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

archetype [#40]

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

diff eq is part of calc 2 iirc

give up on JEE ima beat u anyway

rStellar [#42]

give up on JEE ima beat u anyway

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

adi02 [#39]

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

Thank you man sure

Dreoxx [#41]

diff eq is part of calc 2 iirc

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

archetype [#45]

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

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

Dreoxx [#46]

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

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.

Imagine thinking math is real

i11matic [#48]

Imagine thinking math is real

imagine thinking

archetype [#47]

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.

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

Dreoxx [#50]

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

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

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

archetype [#51]

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

im double majoring in psychology and business so this is a question mark

Dreoxx [#53]

im double majoring in psychology and business so this is a question mark

ooo interesting
yeah i doubt you'd really need to learn this unless you have an interest in mathematics(?) idk

archetype [#54]

ooo interesting
yeah i doubt you'd really need to learn this unless you have an interest in mathematics(?) idk

maybe. i previously was an aspiring engineer since I was pretty far into mathematics, compared to my peers

now I'm not sure if its the field for me, and I feel more passion towards psych, and my skillset towards becoming an actuarian

Dreoxx [#55]

maybe. i previously was an aspiring engineer since I was pretty far into mathematics, compared to my peers

now I'm not sure if its the field for me, and I feel more passion towards psych, and my skillset towards becoming an actuarian

psych is pretty cool. im still in engineering cuz of the moola ngl

archetype [#56]

psych is pretty cool. im still in engineering cuz of the moola ngl

double majoring in psych and business since I follow my passion and the moola jajajaja

Dreoxx [#57]

double majoring in psych and business since I follow my passion and the moola jajajaja

gl o7

archetype [#58]

gl o7

tysm, wishing the best for you too

tanks a million!

ArtemisZ083 [#38]

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

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