Moderate -0.3 This is a straightforward separable variables question requiring integration by parts for the exponential term, followed by substitution of initial conditions and solving for y. While it involves multiple steps (separation, integration by parts, applying boundary condition, numerical evaluation), these are all standard techniques with no novel insight required, making it slightly easier than average.
7 The variables \(x\) and \(y\) are related by the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 x \mathrm { e } ^ { 3 x } } { y ^ { 2 } } .$$
It is given that \(y = 2\) when \(x = 0\). Solve the differential equation and hence find the value of \(y\) when \(x = 0.5\), giving your answer correct to 2 decimal places.
Separate variables correctly and attempt integration on at least one side
M1
Obtain \(\frac{1}{3}y^3\) or equivalent on left-hand side
A1
Use integration by parts on right-hand side (as far as \(axe^{3x} + \int be^{3x}\,dx\))
M1
Obtain or imply \(2xe^{3x} + \int 2e^{3x}\,dx\) or equivalent
A1
Obtain \(2xe^{3x} - \frac{2}{3}e^{3x}\)
A1
Substitute \(x = 0\), \(y = 2\) in an expression containing terms \(Ay^3\), \(Bxe^{3x}\), \(Ce^{3x}\), where \(ABC \neq 0\), and find the value of \(c\)
M1
Obtain \(\frac{1}{3}y^3 = 2xe^{3x} - \frac{2}{3}e^{3x} + \frac{10}{3}\) or equivalent
A1
Substitute \(x = 0.5\) to obtain \(y = 2.44\)
A1
[8]
## Question 7:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Separate variables correctly and attempt integration on at least one side | M1 | |
| Obtain $\frac{1}{3}y^3$ or equivalent on left-hand side | A1 | |
| Use integration by parts on right-hand side (as far as $axe^{3x} + \int be^{3x}\,dx$) | M1 | |
| Obtain or imply $2xe^{3x} + \int 2e^{3x}\,dx$ or equivalent | A1 | |
| Obtain $2xe^{3x} - \frac{2}{3}e^{3x}$ | A1 | |
| Substitute $x = 0$, $y = 2$ in an expression containing terms $Ay^3$, $Bxe^{3x}$, $Ce^{3x}$, where $ABC \neq 0$, and find the value of $c$ | M1 | |
| Obtain $\frac{1}{3}y^3 = 2xe^{3x} - \frac{2}{3}e^{3x} + \frac{10}{3}$ or equivalent | A1 | |
| Substitute $x = 0.5$ to obtain $y = 2.44$ | A1 | [8] |
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7 The variables $x$ and $y$ are related by the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 x \mathrm { e } ^ { 3 x } } { y ^ { 2 } } .$$
It is given that $y = 2$ when $x = 0$. Solve the differential equation and hence find the value of $y$ when $x = 0.5$, giving your answer correct to 2 decimal places.
\hfill \mbox{\textit{CAIE P3 2012 Q7 [8]}}