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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

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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^2}{x} ... 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

Find \( \int \frac{9x+6}{x} \, dx \)

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Answer

To solve this integral, we can simplify the expression inside the integral:

[ \int \frac{9x+6}{x} , dx = \int (9 + \frac{6}{x}) , dx ]

Now, we can integrate term by term:

[ = \int 9 , dx + \int \frac{6}{x} , dx = 9x + 6 \ln |x| + C ]

Step 2

Given that \( y=8 \) at \( x=1 \), solve the differential equation \( \frac{dy}{dx} = \frac{(9x+6)y^2}{x} \)

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Answer

First, rewrite the given equation:

[ \int \frac{1}{y^2} , dy = \int \frac{(9x+6)}{x} , dx ]

Integrating both sides:

[ -\frac{1}{y} = 9 \ln |x| + 6 \ln |x| + C\ ]

Now solve for ( y ):

[ -\frac{1}{y} = 9 \ln |x| + 6 \ln |x| + C \quad \Rightarrow \quad y = -\frac{1}{(9+6) \ln |x| + C} ]

Substituting ( y=8 ) at ( x=1 ):\ [ -\frac{1}{8} = 9 \ln (1) + 6 \ln (1) + C \quad \Rightarrow \quad C = -\frac{1}{8}\ ]

Now substituting back:

[ y^2 = g(x) = 8(3x + 2 \ln x - 3)^2\ ]

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