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7. (a) Express \( \frac{2}{4 - y^2} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 7
Question 8
7. (a) Express \( \frac{2}{4 - y^2} \) in partial fractions. (b) Hence obtain the solution of \[ 2 \cot x \frac{dy}{dx} = (4 - y^2) \] for which \( y = 0 \) at \( x... show full transcript
Worked Solution & Example Answer:7. (a) Express \( \frac{2}{4 - y^2} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 7
Step 1
Express \( \frac{2}{4 - y^2} \) in partial fractions.
Answer
To express ( \frac{2}{4 - y^2} ) in partial fractions, we can factor the denominator:
[ 4 - y^2 = (2 - y)(2 + y) ]
Thus, we can write:
[ \frac{2}{4 - y^2} = \frac{A}{2 - y} + \frac{B}{2 + y} ]
To find ( A ) and ( B ), we multiply through by the denominator ( (2 - y)(2 + y) ):
[ 2 = A(2 + y) + B(2 - y) ]
Expanding gives:
[ 2 = 2A + Ay + 2B - By ]
Rearranging terms:
[ 2 = (2A + 2B) + (A - B)y ]
Now, we equate coefficients:
- For the constant term: ( 2A + 2B = 2 )
- For the coefficient of ( y ): ( A - B = 0 )
From the second equation, we get ( A = B ). Substituting in the first equation gives:
[ 2A + 2A = 2 \implies 4A = 2 \implies A = \frac{1}{2} ]
Thus, ( B = \frac{1}{2} ). Therefore,
[ \frac{2}{4 - y^2} = \frac{1/2}{2 - y} + \frac{1/2}{2 + y} ]
Step 2
Hence obtain the solution of \( 2 \cot x \frac{dy}{dx} = (4 - y^2) \)
Answer
Starting with the equation:
[ 2 \cot x \frac{dy}{dx} = (4 - y^2) ]
We can separate variables:
[ \frac{dy}{4 - y^2} = \frac{2 \cot x}{2} dx \equiv \frac{1}{2 \cot x} dx ]
Integrating both sides gives:
[ \int \frac{dy}{4 - y^2} = \int \frac{1}{2 \cot x} dx ]
The left-hand side integrates to:
[ \frac{1}{4} \ln |2 - y| - \frac{1}{4} \ln |2 + y| = \frac{1}{2} \int \csc x dx ]
Continuing the integration leads us to:
[ y \Rightarrow \frac{1}{4} \ln\left(\frac{2 - y}{2 + y}\right) = \ln(\sec x) + C ]
Using the specified initial condition ( y = 0 ) at ( x = \frac{\pi}{3} ):
At ( x = \frac{\pi}{3} ), ( \sec x = 2 ), substituting gives:
[ 0 = \ln(2) + C \Rightarrow C = -\ln(2) ]
Substituting back:
[ \frac{1}{4} \ln\left(\frac{2 - y}{2 + y}\right) = \ln(\sec x) - \ln(2) ]
This results in:
[ \ln\left(\frac{2 - y}{2 + y}\right) = 4(\ln(\sec x) - \ln(2)) ]
Exponentiating both sides results in:
[ \frac{2 - y}{2 + y} = \frac{\sec^4 x}{16} ]
Finally, rearranging leads us to:
[ \sec^2 x = \frac{2 + y}{2 - y} ]. Thus, the solution in the required form is:
[ \sec^2 x = \frac{8 + 4y}{2 - y} ] or similar equivalents upon further simplifying.