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5. (a) Find \( \int \frac{9x+6}{x} \, dx, \; x > 0 - Edexcel - A-Level Maths Pure - Question 7 - 2010 - Paper 7
Question 7
5. (a) Find \( \int \frac{9x+6}{x} \, dx, \; x > 0. \) (b) Given that \( y = 8 \) at \( x = 1 \), solve the differential equation \( \frac{dy}{dx} = \frac{(9x+6)y^... show full transcript
Worked Solution & Example Answer:5. (a) Find \( \int \frac{9x+6}{x} \, dx, \; x > 0 - Edexcel - A-Level Maths Pure - Question 7 - 2010 - Paper 7
Step 1
Step 2
Given that \( y = 8 \) at \( x = 1 \), solve the differential equation
Answer
We start with the differential equation:
[ \frac{dy}{dx} = \frac{(9x+6)y^2}{x} ]
Rearranging, we have:
[ \frac{1}{y^2} dy = \frac{(9x+6)}{x} dx ]
Integrating both sides, we find:
[ -\frac{1}{y} = 9\ln|x| + 6 \ln|x| + C ]
Thus:
[ -\frac{1}{y} = \frac{9x + 6}{x} + C ]
To find the constant ( C ), we substitute ( y = 8 ) when ( x = 1 ):
[ -\frac{1}{8} = (9 + 6) + C \Longrightarrow C = -\frac{1}{8} - 15 = -\frac{121}{8} ]
The equation becomes:
[ -\frac{1}{y} = \frac{9x + 6}{x} - \frac{121}{8} ]
Now, solving for ( y ):
[ y^2 = g(x) = \left(8(3x + 2 \ln x - 1)\right)^{\frac{1}{2}} ]