Edexcel IAL June 2022 P4 (WMA14/01) Q2 Partial Fractions, Differentiation
a. Express \( \frac{1}{(1+3x)(1-x)} \) in partial fractions.
(3)
(3)
b. Hence find the solution of the differential equation
\( (1+3x)(1-x)\frac{dy}{dx} = \tan y \)
\( -\frac{1}{3} < x \leq \frac{1}{2} \)
For which \( x = \frac{1}{2} \) when \( y = \frac{\pi}{2} \).
Give your answer in the form \( \sin^n y = f(x) \) where n is an integer to be found.
(6)
Solution to Part a: Expression in partial fractions
Key Concepts Used:
- Partial Fractions – Distinct Linear Factors
Step-by-Step Working:
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Set \( \frac{1}{(1+3x)(1-x)} \) identical to \( \frac{A}{1+3x} + \frac{B}{1-x} \).
\( \frac{1}{(1+3x)(1-x)} = \frac{A}{1+3x} + \frac{B}{1-x} \)
-
Let’s find the value of A & B.
Add the two fractions.\( \frac{1}{(1+3x)(1-x)} = \frac{A(1-x) + B(1+3x)}{(1+3x)(1-x)} \)
\( 1 = A(1-x) + B(1+3x) \)
To find B, substitute \( x = 1 \).
\( 1 = A(1-1) + B(1+3 \times 1) \)
\( 1 = 4B \)
\( B = \frac{1}{4} \)
To find A, substitute \( x = -\frac{1}{3} \).
\( 1 = A\left(1 – \left(-\frac{1}{3}\right)\right) + B\left(1 + 3 \times -\frac{1}{3}\right) \)
\( 1 = \frac{4}{3}A \)
\( A = \frac{3}{4} \)
To find B, substitute \( x = 1 \).
\( 1 = A(1-1) + B(1+3 \times 1) \)
\( 1 = 4B \)
\( B = \frac{1}{4} \)
To find A, substitute \( x = -\frac{1}{3} \).
\( 1 = A\left(1 – \left(-\frac{1}{3}\right)\right) + B\left(1 + 3 \times -\frac{1}{3}\right) \)
\( 1 = \frac{4}{3}A \)
\( A = \frac{3}{4} \)
3. Hence,
\( \frac{1}{(1+3x)(1-x)} = \frac{3}{4(1+3x)} + \frac{1}{4(1-x)} \)
\( \frac{1}{(1+3x)(1-x)} = \frac{3}{4(1+3x)} + \frac{1}{4(1-x)} \)
Final Answer:
\( \frac{1}{(1+3x)(1-x)} = \frac{3}{4(1+3x)} + \frac{1}{4(1-x)} \)
Solution to Part b: Solving the Differential Equation
Key Concepts Used:
- Separate Variables: Rearrange to isolate y and x.
- Integrate Both Sides: Use partial fractions from part (a).
- Apply Initial Condition: Find the constant of integration.
Step-by-Step Working:
-
Rewrite the Differential Equation:
\( \frac{dy}{\tan y} = \frac{dx}{(1+3x)(1-x)} \) -
Simplify \( \frac{dy}{\tan y} \):
\( \frac{dy}{\tan y} = \cot y \, dy \)
So, the equation becomes
\( \cot y \, dy = \frac{dx}{(1+3x)(1-x)} \)
where from part (a) we know that
\( \frac{1}{(1+3x)(1-x)} = \frac{3}{4(1+3x)} + \frac{1}{4(1-x)} \)
so, this modifies the equation further
\( \cot y \, dy = \left( \frac{3}{4(1+3x)} + \frac{1}{4(1-x)} \right) dx \) -
Integrate Both Sides:
\( \int \cot y \, dy = \int \left( \frac{3/4}{1+3x} + \frac{1/4}{1-x} \right) dx \)-
Left Side:
\( \int \cot y \, dy = \ln|\sin y| + C_1 \) -
Right Side:
\( \frac{3}{4}.\frac{1}{3} \ln|1+3x| – \frac{1}{4} \ln|1-x| + C_2 \)
\( \frac{1}{4} \ln|1+3x| – \frac{1}{4} \ln|1-x| + C_2 \)
\( \frac{1}{4} \ln|\frac{1+3x}{1-x}| + C_2 \) -
Combine Results:
\[\ln|\sin y| + C_1 = \frac{1}{4} \ln \left| \frac{1 + 3x}{1 – x} \right| + C_2\]\[\ln|\sin y| = \frac{1}{4} \ln \left| \frac{1 + 3x}{1 – x} \right| + C\]
(Remember, the constants C1 and C2 are combined into a single constant C because they both are arbitrary and there is only one constant required in solving differential equation.)
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Left Side:
- Apply Initial Condition \((x = \frac{1}{2}, y = \frac{\pi}{2})\)
\[\ln \left| \sin \frac{\pi}{2} \right| = \frac{1}{4} \ln \left| \frac{1 + 3 \left( \frac{1}{2} \right)}{1 – \frac{1}{2}} \right| + C\]\[\ln |1| = \frac{1}{4} \ln \left| \frac{5}{2} \right| + C\]
\[\ln |1| = \frac{1}{4} \ln \left| \frac{5}{2} \right| + C\]\[0 = \frac{1}{4} \ln |5| + C\]\[C = -\frac{1}{4} \ln |5|\]\[\ln |\sin y| = \frac{1}{4} \ln \left| \frac{1 + 3x}{1 – x} \right| – \frac{1}{4} \ln |5|\]\[\ln |\sin y| = \frac{1}{4} \left( \ln \left| \frac{1 + 3x}{1 – x} \right| – \ln |5|\right)\]\[\ln |\sin y| = \frac{1}{4} \ln \left| \frac{1 + 3x}{5(1 – x)} \right|\]\[\ln |\sin y| = \ln \left| \frac{1 + 3x}{5(1 – x)} \right|^{1/4}\]Cancel log on both sides.\[\sin y = \left( \frac{1 + 3x}{5(1 – x)} \right)^{1/4}\]Raise both sides to the 4th power:\[\sin^4 y = \frac{1 + 3x}{5(1 – x)}\]Final Answer:\[\sin^4 y = \frac{1 + 3x}{5(1 – x)}\]