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3. (a) Express \( \frac{5}{(x-1)(3x+2)} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 5 - 2011 - Paper 6
Question 5
3. (a) Express \( \frac{5}{(x-1)(3x+2)} \) in partial fractions. (b) Hence find \( \int \frac{5}{(x-1)(3x+2)} dx \), where \( x > 1 \). (c) Find the particular sol... show full transcript
Worked Solution & Example Answer:3. (a) Express \( \frac{5}{(x-1)(3x+2)} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 5 - 2011 - Paper 6
Step 1
Express \( \frac{5}{(x-1)(3x+2)} \) in partial fractions.
Answer
To express ( \frac{5}{(x-1)(3x+2)} ) in partial fractions, we assume:
[ \frac{5}{(x-1)(3x+2)} = \frac{A}{x-1} + \frac{B}{3x+2} ]
Multiplying through by the denominator ( (x-1)(3x+2) ) gives:
[ 5 = A(3x+2) + B(x-1) ]
Expanding this, we have:
[ 5 = (3A + B)x + (2A - B) ]
Now we match coefficients:
- For the coefficient of ( x ): ( 3A + B = 0 )
- For the constant term: ( 2A - B = 5 )
From the first equation, we can express ( B ) in terms of ( A ):
[ B = -3A ]
Substituting into the second equation gives:
[ 2A - (-3A) = 5 ]
[ 5A = 5 \Rightarrow A = 1 ]
Thus, substituting back gives:
[ B = -3 ]
Resulting in the partial fractions: [ \frac{5}{(x-1)(3x+2)} = \frac{1}{x-1} - \frac{3}{3x+2} ]
Step 2
Hence find \( \int \frac{5}{(x-1)(3x+2)} dx \), where \( x > 1 \).
Answer
Using the result from part (a):
[ \int \frac{5}{(x-1)(3x+2)} dx = \int \left( \frac{1}{x-1} - \frac{3}{3x+2} \right) dx ]
This separates into two integrals:
[ = \int \frac{1}{x-1} dx - 3 \int \frac{1}{3x+2} dx ]
Calculating these integrals: [ = \ln |x-1| - \ln |3x+2| + C ]
Thus: [ \int \frac{5}{(x-1)(3x+2)} dx = \ln |x-1| - \ln |3x+2| + C ]
Step 3
Find the particular solution of the differential equation \( (x-1)(3x+2) \frac{dy}{dx} = 5y, \quad x > 1, \) for which \( y=8 \) at \( x=2 \).
Answer
We start with:
[ (x-1)(3x+2) \frac{dy}{dx} = 5y ]
Rearranging yields: [ \frac{dy}{y} = \frac{5}{(x-1)(3x+2)} dx ]
Integrating both sides gives:
[ \int \frac{dy}{y} = \int \frac{5}{(x-1)(3x+2)} dx ]
Using the result from part (b):
[ \ln |y| = 5 \left( \ln |x-1| - \ln |3x+2| \right) + C ]
Exponentiating both sides givers: [ y = K \cdot \frac{(x-1)^5}{(3x+2)^5} ]
Now, we use the condition ( y=8 ) at ( x=2 ):
[ 8 = K \cdot \frac{(2-1)^5}{(3(2)+2)^5} \Rightarrow 8 = K \cdot \frac{1}{32} \Rightarrow K = 256 ]
Thus, the particular solution is: [ y = \frac{256(x-1)^5}{(3x+2)^5} ]