# Exact first order differential equations problems

• February 18th 2009, 09:41 AM
Exact first order differential equations problems
I'm totally unsure of where to put this topic - sorry itll probably need moved.(Worried)

Ok I'm doing a further maths A-level right now but I'm also doing an ordinary maths A-level at the same time. This causes a LOT of problems with my further maths, because I have not covered some of the topics in ordinary maths yet.

Anyway i'm having particular problems with Exact 1st order differential equations. It's probably something simple enough that im not getting but I dont understand how:

y + x(dy/dx) = x(cubed)

becomes

(d/dx)(xy) = x(cubed)

In my notes my teacher just wrote (y + x(dy/dx) is exactly the derivative of ?)

What is the question mark.

At the moment we are really rushing through the course, because i started the course late in the year, so doubtless I'll need more help from this forum. (Worried)

Oh and also is there any way to get proper maths notation here. its gonna get hard when it gets to things like integral signs.

Thanks!
• February 18th 2009, 11:05 AM
Chris L T521
Quote:

I'm totally unsure of where to put this topic - sorry itll probably need moved.(Worried)

Ok I'm doing a further maths A-level right now but I'm also doing an ordinary maths A-level at the same time. This causes a LOT of problems with my further maths, because I have not covered some of the topics in ordinary maths yet.

Anyway i'm having particular problems with Exact 1st order differential equations. It's probably something simple enough that im not getting but I dont understand how:

y + x(dy/dx) = x(cubed)

becomes

(d/dx)(xy) = x(cubed)

In my notes my teacher just wrote (y + x(dy/dx) is exactly the derivative of ?)

What is the question mark.

At the moment we are really rushing through the course, because i started the course late in the year, so doubtless I'll need more help from this forum. (Worried)

It may be hard to see at first, but if you treat y as a function of x, $y+x\frac{\,dy}{\,dx}=y\left(1\right)+x\frac{\,dy}{ \,dx}=y\frac{\,d}{\,dx}\!\left(x\right)+x\frac{\,d y}{\,dx}$, which has the form of a product rule application. We can conclude that this is the same as $\frac{\,d}{\,dx}\left(xy\right)$.

Does this make sense?

Quote:

Oh and also is there any way to get proper maths notation here. its gonna get hard when it gets to things like integral signs.

Thanks!
The forum uses LaTeX. You can find out how to use it here. In order to generate the images, the code must be put within  tags. You can also put your mouse cursor over the LaTeX generated images to see the code that was used to produce those images.
• February 18th 2009, 11:37 AM
That does help a lot, and I now get that question - but more complex questions are very tricky to understand . I only get one hour per week of this - so its pretty hectic, and my teacher does not get time to go over much

2xy (dy/dx) + y(squared) = x (cubed)

becomes 4xy (squared) = x^4 +c

My thoughts behind this are: if

: d/dx (x)(y) = y(dx/dx) +x(dy/dx)

Then : d/dx (2xy)(y(squared) ) = y (squared) (1) + (2xy)(dy/dx)

So : (2x)(y(cubed)) = (x^4)/4 + c
And my answer is 8xy(cubed) = x^4 +c

Which... is wrong :( I would assume that my product is wrong yea?

Thanks. :)
• February 18th 2009, 11:50 AM
running-gag
$\frac{\,d(y^2)}{\,dx}=2y \frac{\,dy}{\,dx}$.

$2xy \frac{\,dy}{\,dx} + y^2=x \: 2y \frac{\,dy}{\,dx} + y^2 \left(1\right)=x \: \frac{\,d(y^2)}{\,dx}+y^2 \:\frac{\,dx}{\,dx}$, which has the form of a product rule application. We can conclude that this is the same as $\frac{\,d}{\,dx}\left(xy^2\right)$.

Special tribute to Chris (Rofl)
• February 18th 2009, 12:23 PM
Ah. I didnt know
$\frac{\,d(y^2)}{\,dx}=2y \frac{\,dy}{\,dx}$

wow :)

Gah these are only the most basic of questions of this type and im struggling - my teacher im sure hasnt taught me enough for this (Headbang) I have been working for hours on these first 3 questions of the 18 i have been set. last question before I go - could you explain how

(dy/dx) sin x + y cos x = tan x

becomes y sin x = ln (secx) + c

thanks once again for the help.
• February 18th 2009, 12:40 PM
running-gag
Quote:

Ah. I didnt know
$\frac{\,d(y^2)}{\,dx}=2y \frac{\,dy}{\,dx}$

wow :)

This is a basic rule but if you have never seen this relation I don't know how you can perform your exercises ....

The general rule is $\frac{\,d(y^n)}{\,dx}=ny^{n-1} \frac{\,dy}{\,dx}$

Quote:

Gah these are only the most basic of questions of this type and im struggling - my teacher im sure hasnt taught me enough for this (Headbang) I have been working for hours on these first 3 questions of the 18 i have been set. last question before I go - could you explain how

(dy/dx) sin x + y cos x = tan x

becomes y sin x = ln (secx) + c

thanks once again for the help.

$\frac{d(y\:\sin x)}{dx} = \frac{dy}{dx}\:\sin x + y\:\frac{d(\sin x)}{dx} = \frac{dy}{dx}\:\sin x + y \:\cos x$

And $\tan x = \frac{\sin x}{\cos x} = -\frac{\frac{d(\cos x)}{dx}}{\cos x}$ therefore integration gives $- ln(|\cos x|) + c$
• February 18th 2009, 01:07 PM