Mathematics differential equations

[delsoo]

Homework Statement

hi, all… can anyone help me with this question? i got stucked here? can you figure out which part contains mistake or post the full solution here? thanks in advance!

http://i.imgur.com/RZpquyd.jpg?1

http://imgur.com/RZpquyd&Eoaa0mm#0

Homework Equations

The Attempt at a Solution

 

[Simon Bridge]

Both pics are a bunch of hand-drawn equations and working out.
They are meaningless without the problem they belong to and the reasoning behind them, so I'll have to make some guesses in order to start helping you. In future, please try to describe the problem.

I can see a possible confusion in the line: $$k=\frac{1}{2k}\ln 3$$ ... have you used the same variable for two things here?

There is a red note: $$\frac{dy}{dx}+Py=a$$ ... if you didn't write that, then it may be a hint as to where you went wrong and you should look more closely at the DE you are trying to solve.
(I can't tell if that's supposed to be Py or P_y)

$$ -v\frac{dv}{dk}=h+kv^2$$ ... is what you ended up with.
Assuming that is correct, it tidies up to:$$\frac{dv}{dk}+kv=-\frac{h}{v}$$
Which has form:$$\frac{dv}{dk}+p(k)v=q(v(k))$$... which you appear to have tried to solve via an integrating factor.

I'm guessing that this is what you want help with?

If so then: Compare with "Bernoulli's Equation".
If the writing in red is a hint, it appears to suggest that the initial DE is wrong... so the mistake is off the top of the page.

 

[delsoo]

ps i add in the red note myself ... i am not sure whether can use the 'RED' method or not

 

[delsoo]

Both pics are a bunch of hand-drawn equations and working out.
They are meaningless without the problem they belong to and the reasoning behind them, so I'll have to make some guesses in order to start helping you. In future, please try to describe the problem.

I can see a possible confusion in the line: $$k=\frac{1}{2k}\ln 3$$ ... have you used the same variable for two things here?

There is a red note: $$\frac{dy}{dx}+Py=a$$ ... if you didn't write that, then it may be a hint as to where you went wrong and you should look more closely at the DE you are trying to solve.
(I can't tell if that's supposed to be Py or P_y)

$$ -v\frac{dv}{dk}=h+kv^2$$ ... is what you ended up with.
Assuming that is correct, it tidies up to:$$\frac{dv}{dk}+kv=-\frac{h}{v}$$
Which has form:$$\frac{dv}{dk}+p(k)v=q(v(k))$$... which you appear to have tried to solve via an integrating factor.

I'm guessing that this is what you want help with?

If so then: Compare with "Bernoulli's Equation".
If the writing in red is a hint, it appears to suggest that the initial DE is wrong... so the mistake is off the top of the page.

ps i add in the red note myself ... i am not sure whether can use the 'RED' method or not

so is my working correct? here's the question by the way
http://i.imgur.com/DKtjee4.jpg?1

 

[Simon Bridge]

OK - in general - do not use red pen on your own work.

If you have a DE of form $y'+Py = Q$ where P and Q are functions of x alone, then you can go right to an integrating factor.

But you don't have that form.

You have form: $yy'+xy^2=c$: for c a given constant.

Look up how to solve Bernoulli's equation.

 

[delsoo]

OK - in general - do not use red pen on your own work.

If you have a DE of form $y'+Py = Q$ where P and Q are functions of x alone, then you can go right to an integrating factor.

But you don't have that form.

You have form: $yy'+xy^2=c$: for c a given constant.

Look up how to solve Bernoulli's equation.

thanks for your reply! , i am still confused? would you mind to you show the full working here?

 

[Simon Bridge]

First: Look up how to solve Bernoulli's equation.
If you won't take suggestions I cannot help you.

 

[LCKurtz]

OK - in general - do not use red pen on your own work.

If you have a DE of form $y'+Py = Q$ where P and Q are functions of x alone, then you can go right to an integrating factor.

But you don't have that form.

You have form: $yy'+xy^2=c$: for c a given constant.

Look up how to solve Bernoulli's equation.

thanks for your reply! , i am still confused? would you mind to you show the full working here?

I haven't read the images and don't plan to, but if your problem is to solve something like $yy'+xy^2 = c$, just try substituting $u= y^2,~u'=2yy'$.

 

[delsoo]

First: Look up how to solve Bernoulli's equation.
If you won't take suggestions I cannot help you.

sorry what do u mean by beroullli equation?

 

[delsoo]

http://i.imgur.com/TxlevO8.jpg... well, i stucked this part now... i can't get v=o which the bead stops

 

[delsoo]

First: Look up how to solve Bernoulli's equation.
If you won't take suggestions I cannot help you.

http://i.imgur.com/TxlevO8.jpg... well, i stucked this part now... i can't get v=o which the bead stops

 

[Mark44]

thanks for your reply! , i am still confused? would you mind to you show the full working here?

From the Physics Forums rules, which you agreed to when you joined:

On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.

So no, we won't provide the full solution to your problem.

 

[Simon Bridge]

well, i stucked this part now... i can't get v=o which the bead stops

... did you follow the suggestion?

sorry what do u mean by beroullli equation?

Bernoulli and Google is your friend here:
http://tutorial.math.lamar.edu/Classes/DE/Bernoulli.aspx

You have a relation of form: yy'+xy^2 = c (y=v, x=k, c=-h, right?)

You need to put it in form: y'+py=qyⁿ
... hint: substitute y=1/u and solve for u.

 

[Ray Vickson]

o

thanks for your reply! , i am still confused? would you mind to you show the full working here?

Why are you asking a question? You are asking if you are still confused. If you are confused, just say so, but do not ask whether or not you are---that is what happens when you use a question mark. Now is a good time for you to break a bad habit.

Also: PF rules state specifically that we are not allowed to show you full workings.

 

[delsoo]

o

Why are you asking a question? You are asking if you are still confused. If you are confused, just say so, but do not ask whether or not you are---that is what happens when you use a question mark. Now is a good time for you to break a bad habit.

Also: PF rules state specifically that we are not allowed to show you full workings.

how about partial working?

 

[Ray Vickson]

how about partial working?

Several people have already given you several useful suggestions, but it seems that you are not willing to follow these up. It is up to YOU to do the work; that is the only way you will learn anything.

 

[Simon Bridge]

how about partial working?

... so you didn't even bother to read the link in post #13: that has a partial working.

 

[delsoo]

can you take a look at this question? part b (question 1 ) , my ans is -3/2 ... but the ans given is 2/3... i don't know where's my mistake...
http://imgur.com/ZCrnKA8,bu5bblS
http://imgur.com/ZCrnKA8,bu5bblS#1 thanks in advance!

 

[delsoo]

... so you didn't even bother to read the link in post #13: that has a partial working.

can you take a look at this question? part b (question 1 ) , my ans is -3/2 ... but the ans given is 2/3... i don't know where's my mistake...
http://imgur.com/ZCrnKA8,bu5bblS
http://imgur.com/ZCrnKA8,bu5bblS#1 thanks in advance!

 

[Simon Bridge]

That looks like a different question.
How did you get on with this one?

 

[delsoo]

That looks like a different question.
How did you get on with this one?

can you try to check out where's my mistake please? for the previous question, i realized my mistake finally

 

[Simon Bridge]

Well done.
New question = new thread.