Photo AI
3. (a) Express \( \frac{5}{(x-1)(3x+2)} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 6 - 2011 - Paper 6
Question 6
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 ... 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 6 - 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 it can be represented as:
[ \frac{A}{x-1} + \frac{B}{3x+2} ]
Multiplying through by the denominator ((x-1)(3x+2)", we get:
[ 5 = A(3x+2) + B(x-1) ]
Expanding this, we obtain:
[ 5 = (3A + B)x + (2A - B) ]
Setting up the equations from the coefficients:
- ( 3A + B = 5 )
- ( 2A - B = 0 )
Solving these equations: From (2), we find ( B = 2A ). Substituting into (1):
[ 3A + 2A = 5 \Rightarrow 5A = 5 \Rightarrow A = 1 ]
Then substituting ( A ) back to find ( B ):
[ B = 2(1) = 2 ]
Thus, the partial fraction decomposition is:
[ \frac{5}{(x-1)(3x+2)} = \frac{1}{x-1} + \frac{2}{3x+2} ]
Step 2
Hence find \( \int \frac{5}{(x-1)(3x+2)} \, dx \), where \( x > 1 \).
Answer
Using the partial fraction result:
[ \int \frac{5}{(x-1)(3x+2)} , dx = \int \left( \frac{1}{x-1} + \frac{2}{3x+2} \right) , dx ]
We can split this into two separate integrals:
[ = \int \frac{1}{x-1} , dx + 2 \int \frac{1}{3x+2} , dx ]
Calculating these integrals:
- ( \int \frac{1}{x-1} , dx = \ln |x-1| + C_1 ),
- For the second integral, let ( u = 3x+2 ) then ( dx = \frac{1}{3} du ):
[ \int \frac{1}{3x+2} , dx = \frac{1}{3} \ln |3x+2| + C_2 ]
Therefore, combining both results:
[ \int \frac{5}{(x-1)(3x+2)} , dx = \ln |x-1| + \frac{2}{3} \ln |3x+2| + C ]
Step 3
Find the particular solution of the differential equation \( (x-1)(3x+2) \frac{dy}{dx} = 5y, \quad x > 1, \)
Answer
To find the particular solution, we rewrite the equation:
[ \frac{dy}{dx} = \frac{5y}{(x-1)(3x+2)} ]
This can be solved using separation of variables:
[ \frac{dy}{y} = \frac{5}{(x-1)(3x+2)} , dx ]
Integrating both sides:
[ \int \frac{dy}{y} = 5 \int \frac{1}{(x-1)(3x+2)} , dx ]
We already found the integral of the right side in part (b):
[ \ln |y| = \ln |x-1| + \frac{2}{3} \ln |3x+2| + C ]
Taking exponentials gives us:
[ y = K |x-1| (3x+2)^{\frac{2}{3}} ]
Using the condition ( y = 8 ) when ( x = 2 ):
[ 8 = K |2-1| (3(2)+2)^{\frac{2}{3}} = K \cdot 1 , (8)^{\frac{2}{3}} ]
Thus:
[ K = \frac{8}{4} = 2 ]
Finally, the particular solution is:
[ y = 2 |x-1| (3x+2)^{\frac{2}{3}} ]
In the specified form, we can express it as:
[ y = 2 (x-1) (3x+2)^{\frac{2}{3}} \text{ for } x > 1 ]