Can u=y/x be used to solve x(dy/dx)=y+e^(8x/y)?

[Lchan1]

x(dy/dx)=y+e^(8x/y)
I tried dividing both sides by x
and multiply each term with e^lnx
I kinda solve the problem with x=
but i need to have y=...
please help
and i didnt use to hint of the question of letting u=y/x

 

[rock.freak667]

i didnt use to hint of the question of letting u=y/x

Why did you not use the hint to substitute u=y/x ? Use it and see if it makes things easier.

 

[Lchan1]

I tired and got a wrong answer.. like I take the derivative of my answer and it wouldn't go back to the question.

 

[Hootenanny]

I tired and got a wrong answer.. like I take the derivative of my answer and it wouldn't go back to the question.

Perhaps if you showed your working we could help you out.

 

[Lchan1]

ok
so I did this
x(dy/dx)=y+e^(8x/y)
dy/dx=y/x+(e^(8x/y))/x
dy/dx-y/x=(e^(8x/y))/x sub u=y/x
du/dx-u=(e^8u)/x
multiply both side by e^lnu
d/dx(u-e^lnu)=(e^8u)/x * e^lnu

this is basically what i did. i got an answer late last night and i am in school right now...
couldnt find my work but i got a messy answer.

 

[Hootenanny]

ok
so I did this
x(dy/dx)=y+e^(8x/y)
dy/dx=y/x+(e^(8x/y))/x
dy/dx-y/x=(e^(8x/y))/x sub u=y/x
du/dx-u=(e^8u)/x
multiply both side by e^lnu
d/dx(u-e^lnu)=(e^8u)/x * e^lnu

this is basically what i did. i got an answer late last night and i am in school right now...
couldnt find my work but i got a messy answer.

Are you sure that

$$\frac{dy}{dx}=\frac{du}{dx}$$

 

[Lchan1]

oh i typed it out wrong... sorry... i have x du/dx
cause i used the subsitution dy/dx=xdu/dx..
since u=y/x dy/du=x?
I don't know if it is right... my prof showed us something weird and i am just trying to learn this with my book.

 

[slider142]

oh i typed it out wrong... sorry... i have x du/dx
cause i used the subsitution dy/dx=xdu/dx..
since u=y/x dy/du=x?
I don't know if it is right... my prof showed us something weird and i am just trying to learn this with my book.

If u = y/x, what is du/dx ? You can use the answer to find dy/dx in terms of du/dx, u and x.

 

[Lchan1]

ylnx.
but i don't know how to use it..
my idea was expressing the question in terms of u and x with the subsitution.

 

[Hootenanny]

oh i typed it out wrong... sorry... i have x du/dx
cause i used the subsitution dy/dx=xdu/dx..
since u=y/x dy/du=x?
I don't know if it is right... my prof showed us something weird and i am just trying to learn this with my book.

That's not quite right:

$$u = \frac{y}{x} \Rightarrow \frac{du}{dx} = \frac{d}{dx}\frac{y}{x}$$

Note that here y=y(x), that is y is a function of x. So you need to differentiate a product (or quotient) of two functions of x. Do you know how to evaluate the derivative of this form?

HINT:

Spoiler

Use the product/quotient rule and the chain rule

 

[Lchan1]

I am confused now. If i do du/dx = xdy/dx-y/x^2
I will get an equation with 3 different variables
let me show you what i did.
x(dy/dx)=y+e^(8x/y)
dy/dx=y/x+(e^(8x/y))/x
dy/dx-y/x=(e^(8x/y))/x sub u=y/x

since dy/dx= (dy/du)(du/dx) dy/du=x and du/dx=(xdy/dx-y)/x^2
multiply to replace dy/dx. i get (xdy/dx-y)/x-u=(e^8u)/x

 

[slider142]

I am confused now. If i do du/dx = xdy/dx-y/x^2...

Note that since u = y/x, we have y = ux, so your expression becomes u' = (y' - u)/x, so we can now get y' as a function of u', u, and x.

 

[Lchan1]

I got an answer but the want it interms of y= no way i can separate it..
i got y=u^2+1/8e^8u
put u=y/x

 

[The Dagda]

I got an answer but the want it interms of y= no way i can separate it..
i got y=u^2+1/8e^8u
put u=y/x

I may be being dense here but as someone else mentioned can you use the quotient rule on that?

$$f(x) = \frac{g(x)}{h(x)}$$

$$\frac{d}{dx}f(x) = f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$$

And as said the chain rule, have you studied this yet?

 

[Lchan1]

well i know the quotient rule for derivatives...
i or chain rule for derivatives...
other than that.. I don't know any other chain rule...

 

[The Dagda]

Ok have you tried plugging it in as g(x) and h(x) what do you get? Can you then use the chain rule also?

Sorry I'm not really paying attention to this thread, doing 8 things at once. Hope that helps...

 

[Lchan1]

well i used the chain rule to find u' and i am stuck from there
i am ask to solve for the dfferential equation. i try to integral both sides and got an answer.
but the answer is not solve for y=

 

[Hootenanny]

I got an answer but the want it interms of y= no way i can separate it..
i got y=u^2+1/8e^8u

That is not correct. Perhaps if you posted your working we could point out where you have made an error.

 

[Lchan1]

k I just found out i type my question wrong here originally. but this is what i did and got another answer late last night.
u'=(y'-u)/x
y'=u+e^8u
u'=(e^8u)/x
ln(x)=-1/8e^8u

ln(ln(x))=-u
y=-ln(ln(x))/x
the 1/8 is gone which makes my answer weird...
since you will get the same answer for different e^ whatever x...

 

[Hootenanny]

Can I just clear one thing up first:

x(dy/dx)=y+e^(8x/y)

Did you mean e^{8y/x}?

 

[Lchan1]

Can I just clear one thing up first:

Did you mean e^{8y/x}?

the orginally question should have been
xy'=y+xe^{8y/x}

which gives y'=y/x+e^{8y/x}

 

[Lchan1]

you totally found another mistake i had typing my question..
I am so sorry and thanks for all these help..
I got the topic a lot better and i could do some of the same type in the book.

 

[Hootenanny]

the orginally question should have been
xy'=y+xe^{8y/x}

which gives y'=y/x+e^{8y/x}

you totally found another mistake i had typing my question..
I am so sorry and thanks for all these help.

Not a problem, no need to apologise. Now, back to your question:

k I just found out i type my question wrong here originally. but this is what i did and got another answer late last night.
u'=(y'-u)/x
y'=u+e^8u
u'=(e^8u)/x
ln(x)=-1/8e^8u

You can't simply solve u' = (y'-u)/x. You need to substitute this back into the original ODE.

 

[Lchan1]

I did this too... Let me show you my work for that..
Seriously i tried this question for at least 3 hours. and i did every possible thing i could do.
solve for y' in the u' equation
y'=u'x+u
u'x+u=u+e^8u
1/e^8u du = 1/x dx
lnx= -1/8 e^-8u

it's the same problem i ran into again.

 

[Hootenanny]

the orginally question should have been
xy'=y+xe^{8y/x}

Are you sure that this is the probably, exactly as given? Specifically, are you sure that the exponential has a coefficient of x?

 

[Lchan1]

Yes...
xy'=y+xe^{8y/x)
so when you divided both sides by x...

 

[Lchan1]

x y' = y + xe^((8 y)/x)
that's the question I copy off from my online practice.

 

[Hootenanny]

x y' = y + xe^((8 y)/x)
that's the question I copy off from my online practice.

I did this too... Let me show you my work for that..
Seriously i tried this question for at least 3 hours. and i did every possible thing i could do.
solve for y' in the u' equation
y'=u'x+u
u'x+u=u+e^8u
1/e^8u du = 1/x dx
lnx= -1/8 e^-8u

it's the same problem i ran into again.

Then in that case, your solution is correct:

$$\ln\left(\frac{1}{x^8}\right) = e^{-8\frac{y}{x}}$$

All you need to do now is take the logarithm and re-arrange. However, you must be careful to define the domain of your solution.

 

[Lchan1]

oh i finally finished... Thank you so much
I got y = ln(ln(1/x^8))(x)/-8

and domain is x cannot be zero.

 

[Hootenanny]

and domain is x cannot be zero.

Almost, but note quite. Note that the argument of a logarithm cannot be negative and that the logatithm of a number less than one is negative.

 

[Lchan1]

oh. e... hahahaa thanks