Moderate -0.3 This is a straightforward separable variables question requiring standard algebraic manipulation (splitting the fraction into y + 1/y²), integration of simple terms, and applying an initial condition. While it requires multiple steps, each technique is routine for P3 level with no conceptual challenges or novel insights needed.
6 Given that \(y = 1\) when \(x = 0\), solve the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y ^ { 3 } + 1 } { y ^ { 2 } }$$
obtaining an expression for \(y\) in terms of \(x\).
Obtain terms \(\frac{1}{3}\ln(y^3+1)\) and \(x\), or equivalent
A1 + A1
Evaluate a constant or use limits \(x = 0\), \(y = 1\) with a solution containing terms \(k\ln(y^3+1)\) and \(x\), or equivalent
M1
Obtain any correct form of solution e.g. \(\frac{1}{3}\ln(y^3+1) = x + \frac{1}{3}\ln 2\)
A1\(\sqrt{}\)
Rearrange and obtain \(y = (2e^{3x}-1)^{\frac{1}{3}}\), or equivalent
A1
Total: 6
[f.t. is on \(k \neq 0\)]
## Question 6:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Separate variables and attempt to integrate | M1 | |
| Obtain terms $\frac{1}{3}\ln(y^3+1)$ and $x$, or equivalent | A1 + A1 | |
| Evaluate a constant or use limits $x = 0$, $y = 1$ with a solution containing terms $k\ln(y^3+1)$ and $x$, or equivalent | M1 | |
| Obtain any correct form of solution e.g. $\frac{1}{3}\ln(y^3+1) = x + \frac{1}{3}\ln 2$ | A1$\sqrt{}$ | |
| Rearrange and obtain $y = (2e^{3x}-1)^{\frac{1}{3}}$, or equivalent | A1 | **Total: 6** |
| [f.t. is on $k \neq 0$] | | |
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6 Given that $y = 1$ when $x = 0$, solve the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y ^ { 3 } + 1 } { y ^ { 2 } }$$
obtaining an expression for $y$ in terms of $x$.
\hfill \mbox{\textit{CAIE P3 2004 Q6 [6]}}