differentiating partial fraction

• Apr 8th 2010, 01:49 PM
sigma1
differentiating partial fraction
i have been given a function to be expressed in partial fractions then differentiate it.
$\displaystyle \frac{4x^2+3x-2} {(u+1)(2u+3)}$

am stuck at the point where i cant eliminate one of the variables.

using $\displaystyle \frac{4x^2+3x-2} {(u+1)(2u+3)}$

$\displaystyle 4x^2+3x-2 ={A}/(u+1)+{B}/(2u+3)$
$\displaystyle 4x^2+3x-2 = A(2U+3)+B(U+1)$

eliminating B by letting U = -1
therefore A =-1 (assuming i did it correctly)
i cant seem to eliminate A to get a value for B. am i doing something wrong.
• Apr 8th 2010, 05:18 PM
dwsmith
Didn't work nevermind. This problem is killing me.
• Apr 8th 2010, 06:48 PM
sigma1
sorry i got a bit carried away with the variables. the question actually states.

integrate the following with respects to U by first expressing as partial fractions.

[tex]

4u^2+3u-2 / (u+1)(2u+3)

could you show me how its done without using matrices..
• Apr 8th 2010, 06:59 PM
dwsmith
Are all the numbers correct in the original problem?
• Apr 8th 2010, 07:20 PM
dwsmith
This linear system is inconsistent so I think there is something missing in the problem.
• Apr 8th 2010, 07:24 PM
sigma1
yes they are
• Apr 8th 2010, 07:38 PM
dwsmith
Nevermind
• Apr 8th 2010, 07:49 PM
dwsmith
The problem we are facing is that we have a $\displaystyle 4x^2$ but no $\displaystyle x^2$ terms with our As and Bs in the partial fraction equation systems.

That means we have $\displaystyle 0=4$ but $\displaystyle 0 \neq 4$
• Apr 8th 2010, 08:23 PM
sigma1
i think i may have go this one
$\displaystyle \frac{4u^2+3u-2} {(u+1)(2u+3)}$

$\displaystyle 4u^2+3u-2=A(2u+3)+B(u+1)$

letting u -1 i solve the equation by eliminating B to get a value for A which is -1 ..... A= -1

letting u -1.5 i solve the equation by eliminating A to get a value for B which is -5 ..... B= -5

partial fraction =
$\displaystyle \frac{-1} {(u+1)}$ +
$\displaystyle \frac{-5} {(2u+3)}$

now all i have to do is diffrenciate that. think the quotient rule is needed there.. any help?
• Apr 8th 2010, 08:26 PM
dwsmith
$\displaystyle -(2u+3)-5(u+1)\neq 4u^2+3u-2$
• Apr 8th 2010, 08:29 PM
dwsmith
I don't think this can be decomposed with partial fractions since it is inconsistent.
• Apr 8th 2010, 08:39 PM
sigma1
wow i see what you mean...thanks for all your help though.. i appreciate it.
• Apr 8th 2010, 08:47 PM
harish21
Quote:

i think i may have go this one
$\displaystyle \frac{4u^2+3u-2} {(u+1)(2u+3)}$

$\displaystyle 4u^2+3u-2=A(2u+3)+B(u+1)$

letting u -1 i solve the equation by eliminating B to get a value for A which is -1 ..... A= -1

letting u -1.5 i solve the equation by eliminating A to get a value for B which is -5 ..... B= -5

partial fraction =
$\displaystyle \frac{-1} {(u+1)}$ +
$\displaystyle \frac{-5} {(2u+3)}$

now all i have to do is diffrenciate that. think the quotient rule is needed there.. any help?

Your partial fraction is not correct. This is what you should be getting.

$\displaystyle \frac{4u^2+3u-2}{(u+1)(2u+3)} = 2-\frac{5}{2u+3} - \frac{1}{u+1}$

Differentiating this gives:

$\displaystyle \frac{dy}{du} = 0 - \frac {-10}{(2u+3)^2} - \frac{-1}{(u+1)^2} = \frac {10}{(2u+3)^2} + \frac{1}{(u+1)^2}$
• Apr 9th 2010, 03:53 AM
sigma1
Quote:

Your partial fraction is not correct. This is what you should be getting.

$\displaystyle \frac{4u^2+3u-2}{(u+1)(2u+3)} = 2-\frac{5}{2u+3} - \frac{1}{u+1}$

Differentiating this gives:

$\displaystyle \frac{dy}{du} = 0 - \frac {-10}{(2u+3)^2} - \frac{-1}{(u+1)^2} = \frac {10}{(2u+3)^2} + \frac{1}{(u+1)^2}$

if its not too much trouble could you please help me a bit and explain how you arrived at you answer? thanks
• Apr 9th 2010, 09:52 AM
harish21
Quote:

if its not too much trouble could you please help me a bit and explain how you arrived at you answer? thanks

Yes.

Your question is telling you to decompose the function into partial fractions.

$\displaystyle \frac{4u^2+3u-2} {(u+1)(2u+3)}$

Notice that both numerator and denominators are polynomials of degree 2. SO you can decompose the given function by:

$\displaystyle y = \frac{4u^2+3u-2} {(u+1)(2u+3)} = {A} + \frac{B}{u+1} + \frac{C}{2u+3}$

This gives

$\displaystyle {4u^2+3u-2} = A(u+1)(2u+3) + B(2u+3) + C(u+1)$

Now to find A, B, and C,

first let u=1, this gives B = -1

then let $\displaystyle u = \frac{-3}{2}$, this gives C = -5

and then let u = 1, this gives A=2.

So your function is decomposed into partial fractions as:

$\displaystyle \frac{4u^2+3u-2} {(u+1)(2u+3)} = {2} - \frac{1}{u+1} - \frac{5}{2u+3}$

Now,you are supposed to find the derivative of the right hand side of the equation, which is the function decomposed into partial fractions.

You have posted this question under "Calculus" section of this forum. So, I am assuming you have enough knowledge of derivatives.

However, I did the derivative finding part yesterday in my previous post (using the quotient rule).

Remember: Derivative of a constant (which is ,2) = 0

Is it clear now?