Moderate -0.3 This is a straightforward separable variables question with standard integration techniques. Part (a) is routine algebraic manipulation and integration, while part (b) requires separation of variables and applying an initial condition—both are standard C4 procedures with no conceptual challenges or novel insights required.
5. (a) Find \(\int \frac { 9 x + 6 } { x } \mathrm {~d} x , x > 0\).
(b) Given that \(y = 8\) at \(x = 1\), solve the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( 9 x + 6 ) y ^ { \frac { 1 } { 3 } } } { x }$$
giving your answer in the form \(y ^ { 2 } = \mathrm { g } ( x )\).
\section*{LU}
\(\int \frac{1}{y^{\frac{1}{3}}} \, dy = \int \frac{9x+6}{x} \, dx\) then \(\int y^{-\frac{1}{3}} \, dy = \int \frac{9x+6}{x} \, dx\) then \(\frac{y^{\frac{2}{3}}}{\frac{2}{3}} = 9x + 6\ln x (+C)\) then \(\frac{3}{2}y^{\frac{1}{3}} = 9x + 6\ln x (+C)\)
Integral signs not necessary B1; \(\pm ky^{\frac{1}{3}}\) = their (a) M1; A1fft;
When \(y = 8, x = 1\): \(\frac{3}{2}8^{\frac{1}{3}} = 9 + 6\ln 1 + C\) then \(C = -3\) then \(y^{\frac{1}{3}} = \frac{2}{3}(9x + 6\ln x - 3)\) then \(y^2 = (6x + 4\ln x - 2)^2\) (or \(= 8(3x + 2\ln x - 1)^2\))
M1; A1; A1
(6 marks)
Total: [8]
(a) | $\int \frac{9x+6}{x} \, dx = \int \left(9 + \frac{6}{x}\right) \, dx = 9x + 6\ln x (+C)$ | M1; A1 | (2 marks)
(b) | $\int \frac{1}{y^{\frac{1}{3}}} \, dy = \int \frac{9x+6}{x} \, dx$ then $\int y^{-\frac{1}{3}} \, dy = \int \frac{9x+6}{x} \, dx$ then $\frac{y^{\frac{2}{3}}}{\frac{2}{3}} = 9x + 6\ln x (+C)$ then $\frac{3}{2}y^{\frac{1}{3}} = 9x + 6\ln x (+C)$ | Integral signs not necessary B1; $\pm ky^{\frac{1}{3}}$ = their (a) M1; A1fft;
When $y = 8, x = 1$: $\frac{3}{2}8^{\frac{1}{3}} = 9 + 6\ln 1 + C$ then $C = -3$ then $y^{\frac{1}{3}} = \frac{2}{3}(9x + 6\ln x - 3)$ then $y^2 = (6x + 4\ln x - 2)^2$ (or $= 8(3x + 2\ln x - 1)^2$) | M1; A1; A1 | (6 marks)
**Total: [8]**
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5. (a) Find $\int \frac { 9 x + 6 } { x } \mathrm {~d} x , x > 0$.\\
(b) Given that $y = 8$ at $x = 1$, solve the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( 9 x + 6 ) y ^ { \frac { 1 } { 3 } } } { x }$$
giving your answer in the form $y ^ { 2 } = \mathrm { g } ( x )$.
\section*{LU}
\hfill \mbox{\textit{Edexcel C4 2010 Q5 [8]}}