MORE Calc help...please

[CrazyThomas 0]

So, there's this equation.....needs to
be solved for f(x)

eqn is:

dx/dy = x + y

I'm thinking the solution is :

f(x) = xy

Since the product rule says
f(x)'g(x) +f(x)g(x)'
So, unless implicit differentiation is hosing me up, wouldn't f (x) = xy product rule out to be x + y ?

Or is implicit differentiation used a different way?


Thanks,
Thomas

[flyinblind 0]

I'm not sure I understand exactly what you've got or else I would help. The problem I see with what you proposed is that the product rule applies to differentiation with respect to one variable. So in the case of xy when you differentiate with respect to x you would get just y. And vice versa if you differentitate with respect to y you would get just x. But that in know way get's the two to be added together.

However, if you had something like f(x) = 2x^2 + xy, then differentiating with respect to x would give you x + y.

Like I said, I'm not terribly sure what you've got. And Calc 3 was two semesters ago so I'm a bit rusty.


ymmv...


cheers,


timmah

cheers and blue skies,

timmah

[brits17 0]

Its been a while but...

A good way to remember the product rule for differentiation is "the first times the derivative of the second plus the second times the derivative of the first.''

Which is what you stated with f(x)'g(x) + g(x)'f(x). If the derivative of x is 1x^0 and the derivative of y is 1y^0, then dx/dy would come out to be 1x + 1y = x+y.

I thought it out backwards, but came up with your same conclusion. Again, its been a long while so I'm just throwing stuff out there...

_______________________
aerialkinetics.com

[pleifer 0]

that is not product rule

and I am trying to find for implicit diff

so far, to solve for f'(x) i think we need an x value??

edited to add- product rule would be f(x)=xy

_________________________________________
The Angel of Duh has spoke

[pleifer 0]

I think the anwser for implicit diff is

dy/dy= x+y

is f(x)=1

tell me if I am wrong, or if you need me to explain how I got this

_________________________________________
The Angel of Duh has spoke

[flyinblind 0]

I'm pretty sure that's wrong. If you differentiate that with respect to x you get 0. So that's not right.

Here's a good link about implicit differentiation, but you kind of need implicit integration to work backward.

http://archives.math.utk.edu/visual.calculus/3/implicit.7/

cheers and blue skies,

timmah

[VivaHeadDown 0]

I have gotten a different answer everytime, and I'm 99% sure their all wrong

I give up

You put dx/dy in your original statement, did you mean dy/dx?

Don't Confuse Me With My Own Words

[JackC 0]

Solve y'-y = x

Using the "integrating factor" method for first order linear differential equations of the form y'+P(x)y=Q(x) where the integrating factor is : R(x)=exp( integral{ P(x) dx })

In this case P(x)=-1 so R(x)=exp(-x)

Multiply through by R(x) to give: exp(-x) y' - y exp(-x) = x exp(-x)

The LHS is just d/dx{ y exp(-x)} which we substitute to give:

d/dx{y exp(-x)} = x exp(-x)

Integrate both sides with respect to x to give

y exp(-x) = (-x - 1) exp(-x) + C

where C is the constant of integration. Rearange to give:

y=C exp(x) - x - 1

We can check that this is correct by substitution into the original expression.

So y'-y = (C exp(x) - 1) - (C exp(x) - x -1) = x as required.

[falxori 0]

your whole representation is wrong.
if: f(x,y)= xy
then the partial differentiations are:
df(x,y)/dx =y
df(x,y)/dy=x

what you wrote "dx/dy" has no meaning because x is not a function of y, they are both variables of f(x,y)
the product rule would work if y is also a function of x.

O

"Carpe diem, quam minimum credula postero."

[kallend 2,146]

Quote

Solve y'-y = x

Using the "integrating factor" method for first order linear differential equations of the form y'+P(x)y=Q(x) where the integrating factor is : R(x)=exp( integral{ P(x) dx })

In this case P(x)=-1 so R(x)=exp(-x)

Multiply through by R(x) to give: exp(-x) y' - y exp(-x) = x exp(-x)

The LHS is just d/dx{ y exp(-x)} which we substitute to give:

d/dx{y exp(-x)} = x exp(-x)

Integrate both sides with respect to x to give

y exp(-x) = (-x - 1) exp(-x) + C

where C is the constant of integration. Rearange to give:

y=C exp(x) - x - 1

We can check that this is correct by substitution into the original expression.

So y'-y = (C exp(x) - 1) - (C exp(x) - x -1) = x as required.

Unless he wrote in incorrectly in the first place
"dx/dy = x + y"

he has 'dx/dy', NOT 'dy/dx' in there, so the equation isn't y'-y = x, it's y' = 1/(x+y). You can transform, but you need to back-transform your solution.

...

The only sure way to survive a canopy collision is not to have one.

[brits17 0]

Quote

IYou put dx/dy in your original statement, did you mean dy/dx?

Thats what I'm thinkin...

_______________________
aerialkinetics.com

[CrazyThomas 0]

Quote

I have gotten a different answer everytime, and I'm 99% sure their all wrong

I give up

You put dx/dy in your original statement, did you mean dy/dx?

Yes. I wrote it down wrong. Sorry about that. That does make a big bit of difference.
The problem is

dy/dx = x + y

solve for f (x)

In which case it looks like the statement
f (x) = (x^2)/2 +xy
should work as a solution.

If I am thinking clearly, dy/dx means the derivative of y, with respect to x. in which case, y is left alone, and the xy part differentiates into 1y.

Thanks for the help.


Thomas

[CrazyThomas 0]

Yes, like the previous post states, I wrote the problem down incorrectly.
Sorry about that. It helps to have a correctly written problem when trying to offer help.


Thomas

[JackC 0]

Quote

The problem is

dy/dx = x + y

solve for f (x)

I assume you mean solve y'=x+y for y=f(x). If y isn't a function of x the expression is x+y=0

Quote

In which case it looks like the statement
f (x) = (x^2)/2 +xy
should work as a solution.

No. What you have there is f(x) = x^2/2+x f(x) which isn't a solution. Substitute it into the original expression and it doesn't work.

Quote

If I am thinking clearly, dy/dx means the derivative of y, with respect to x.

Yep, for regular (1 dimensional) differentiation thats right.

Quote

in which case, y is left alone, and the xy part differentiates into 1y.

Nope, that's partial differentiation. To differentiate xy with respect to x where y=f(x) requires implicit differentiation, this is:

d/dx{xy}= y + x dy/dx

To solve the original differential equation, you need to use the integrating factor method shown in my previous post, or do what most folks do, and guess a trial solution and substitute it into the differential equation, then see how it works.

For example, solve y'-y=x.

First solve y'-y=0 (solutions are known as the complementary function) by guessing y=C exp(ax) so that y'=a C exp(ax) which gives

a C exp(ax) - C exp(ax) = 0

Which is correct only if a=1, so we know that the complementary function is y=C exp(x).

Next we solve for the particular integral y'-y=x by guessing y=ax so that y'=a giving

a-ax=x. This doesn't work so we try again.

Let y=ax+b so that y'=a giving

a-ax-b=x.

This only works if a=b=-1

So we know that the particular integral is y=-x-1

Put them together and we have the general solution which is:

y=C exp(x) - x - 1, the same as using the integrating factor method.

Solving differential equations is a bit more tricky than just looking for the reverse of differentiation.

[JeffD 0]

I think you guys looked at this one too critically.

dy/dx = x + y , now multiply each side by dx so you have

dy = xdx + ydx , now integrate
y = x^2/2 + yx , now solve for y

y - yx = (x^2)/2
y(1-x)=(x^2)/2
y = ((x^2)/2)/(1-x),

• Quote

[CrazyThomas 0]

Quote

I think you guys looked at this one too critically.

dy/dx = x + y , now multiply each side by dx so you have

dy = xdx + ydx , now integrate
y = x^2/2 + yx , now solve for y

y - yx = (x^2)/2
y(1-x)=(x^2)/2
y = ((x^2)/2)/(1-x),

hmm....interesting.
y = ((x^2)/2)/(1-x)

doesn't this simplify to:
y = (x^2 * (1-x)) / 2 ?

in which case, you could FOIL it out to be
y = (x^2 - x^3 ) / 2 ?

So, y' = x - 1.5x^2

Checking original eqn, :
y' - y = x

[x-1.5x^2] - [(x^2 - x^3) /2 ] = x

This does not work, as there is no way to get rid of the x^3 term.

The integration factor method is what I've been seeing in class, and seems to be the correct method for this.
especially when I see that exp(x) term pop up.

Thanks again for the help.

Thomas

[VivaHeadDown 0]

no, it simplifies to

y = (x^2) / (2*(1-x))

Don't Confuse Me With My Own Words