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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 \(... 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 first recognize that ( 4 - y^2 ) can be factored as ( (2 - y)(2 + y) ). Thus, we can write the expression as:
[ \frac{2}{4 - y^2} = \frac{2}{(2 - y)(2 + y)} = \frac{A}{2 - y} + \frac{B}{2 + y} ]
Multiplying through by the denominator, we have:
[ 2 = A(2 + y) + B(2 - y) ]
By choosing convenient values for ( y ), we can solve for ( A ) and ( B ):
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Let ( y = 2 ): [ 2 = A(4) + B(0) \Rightarrow A = \frac{1}{2} ]
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Let ( y = -2 ): [ 2 = A(0) + B(4) \Rightarrow B = \frac{1}{2} ]
So we have:
[ \frac{2}{4 - y^2} = \frac{1/2}{2 - y} + \frac{1/2}{2 + y} ]
This completes the partial fraction decomposition.
Step 2
Hence obtain the solution of \( 2 \cot x \frac{dy}{dx} = (4 - y^2) \).
Answer
Starting with the rewritten equation:
[ 2 \cot x \frac{dy}{dx} = (4 - y^2) ]
We can separate variables by rearranging terms:
[ \frac{dy}{4 - y^2} = \frac{1}{2 \cot x} dx ]
Now, integrating both sides:
[ \int \frac{dy}{4 - y^2} = \int \frac{1}{2 \cot x} dx ]
The left side can be integrated as:
[ \int \frac{1}{4 - y^2} dy = \frac{1}{4} \ln |4 - y^2| + C_1 ]
The right side leads to:
[ \int \frac{1}{2 \cot x} dx = \frac{1}{2} \int \tan x , dx = \frac{1}{2} (-\ln | \cos x |) + C_2 ]
Hence, equating gives:
[ \frac{1}{4} \ln |4 - y^2| = -\frac{1}{2} \ln | \cos x | + C ]
To find the constant of integration, we use the condition ( y = 0 ) at ( x = \frac{\pi}{3} ):
[ 0 = -\frac{1}{2} \ln | \cos(\frac{\pi}{3}) | + C \Rightarrow 0 = -\frac{1}{2} \ln (\frac{1}{2}) + C \Rightarrow C = \frac{1}{2} \ln (2) ]
Thus, substituting back yields:
[ \frac{1}{4} \ln |4 - y^2| = -\frac{1}{2} \ln | \cos x | + \frac{1}{2} \ln (2) ]
Rearranging gives:
[ \ln |4 - y^2| = -2 \ln | \cos x | + 2 \ln (2) = \ln(\frac{2^2}{\cos^2 x}) = \ln(\frac{4}{\cos^2 x}) ]
Exponentiating both sides:
[ |4 - y^2| = \frac{4}{\cos^2 x} \Rightarrow 4 - y^2 = \frac{4}{\cos^2 x} ]
Rearranging results in:
[ \sec^2 x = \frac{2 + y}{2 - y} ]
Thus, we have shown that ( \sec^2 x = g(y) ).