Photo AI
7. (a) Express \( \frac{2}{4 - y^2} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 1 - 2008 - Paper 7
Question 1
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 \( y = 0 \) a... show full transcript
Worked Solution & Example Answer:7. (a) Express \( \frac{2}{4 - y^2} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 1 - 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 recognize that ( 4 - y^2 ) can be factored as ( (2 - y)(2 + y) ). Therefore, we can set up the equation: [ \frac{2}{(2 - y)(2 + y)} = \frac{A}{2 - y} + \frac{B}{2 + y} ] Multiplying through by the denominator ( (2 - y)(2 + y) ), we get: [ 2 = A(2 + y) + B(2 - y) ] Expanding the right-hand side, we have: [ 2 = 2A + Ay + 2B - By ] Grouping terms gives: [ 2 = (2A + 2B) + (A - B)y ] To satisfy this equation for all values of ( y ), the coefficients must equal zero. Therefore, we equate:
- ( 2A + 2B = 2 )
- ( A - B = 0 )
From equation (2), we can express ( A ) in terms of ( B ): ( A = B ).
Substituting this into equation (1):
[ 2A + 2A = 2 \implies 4A = 2 \implies A = \frac{1}{2}
] Thus, we find ( A = B = \frac{1}{2} ). The final expression in partial fractions is: [ \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
To solve the differential equation ( 2 \cot x \frac{dy}{dx} = (4 - y^2) ), we start by separating variables:
[
\frac{dy}{4 - y^2} = \frac{1}{2 \cot x} dx
]
Next, we integrate both sides:
[
\int \frac{dy}{4 - y^2} = \int \frac{1}{2 \cot x} dx
]
The left side integrates to:
[
\frac{1}{2} \ln |2 - y| - \frac{1}{2} \ln |2 + y| = \frac{1}{2} \ln\left(\frac{2 - y}{2 + y}\right) + C
]
For the right-hand side:
[
\int \frac{dx}{2 \cot x} = \frac{1}{2} \int \tan x , dx = -\frac{1}{2} \ln |\cos x| + C'
]
Setting both sides equal gives:
[
\frac{1}{2} \ln\left(\frac{2 - y}{2 + y}\right) = -\frac{1}{2} \ln |\cos x| + C
]
Now, at ( x = \frac{\pi}{3} ), ( y = 0 ):
[
\frac{2 - 0}{2 + 0} = \frac{2}{2} = 1
]
This means:
[ 0 = -\frac{1}{2} \ln |\cos(\frac{\pi}{3})| + C \implies C = \frac{1}{2} \ln 2 ]
Substituting ( C ) back gives:
[ \ln\left(\frac{2 - y}{2 + y}\right) = -\ln |\cos x| + \ln 2 ]
Combining the logarithms results in:
[ \frac{2 - y}{2 + y} = k \cdot \cos x \quad \text{(where ( k ) is a constant)} ]
Rearranging leads to:
[ 2 - y = k \cos x (2 + y) \quad \Rightarrow \quad y(1 + k \cos x) = 2 - 2k \cos x
]
Thus:
[ y = \frac{2 - 2k \cos x}{1 + k \cos x}
]
Finally, making ( y = 0 ) will help in finding the specific form for ( g(y) ). Hence, it simplifies to:
[ \sec^2 x = \frac{2 + y}{2 - y}
] in the required form.