CAIE P3 2021 June — Question 7 7 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2021
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFirst order differential equations (integrating factor)
TypeDeriving the differential equation
DifficultyStandard +0.3 This is a straightforward first-order separable differential equation question. Students must translate 'gradient proportional to' into dy/dx = k·y/√(x+1), separate variables, integrate both sides (standard integrals), then use two given points to find constants. While it requires multiple steps, each is routine and the question clearly signposts the method ('setting up and solving'). Slightly easier than average due to the explicit guidance and standard technique.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)

7 A curve is such that the gradient at a general point with coordinates \(( x , y )\) is proportional to \(\frac { y } { \sqrt { x + 1 } }\). The curve passes through the points with coordinates \(( 0,1 )\) and \(( 3 , e )\). By setting up and solving a differential equation, find the equation of the curve, expressing \(y\) in terms of \(x\).

Question 7:
AnswerMarks Guidance
AnswerMark Guidance
State equation \(\frac{dy}{dx} = k\frac{y}{\sqrt{x+1}}\)B1 OE. Must be a differential equation.
Separate variables correctly for *their* differential equation and integrate at least one side\*M1 \(\int\frac{1}{y}dy = \int\frac{k}{\sqrt{x+1}}dx\)
Obtain \(\ln y\)A1 Allow M1A1A1 if they have assumed \(k=1\) or are working with an incorrect value for \(k\)
Obtain \(2[k]\sqrt{x+1}\)A1
Use \((0,1)\) and \((3,e)\) in an expression containing \(\ln y\) and \(\sqrt{x+1}\) and a constant of integration to determine \(k\) and/or a constant of integration \(c\) (or use \((0,1)\), \((3,e)\) and \((x,y)\) as limits on definite integrals)DM1 If remove logs before finding the constant of integration then the constant must be of the correct form.
Obtain \(k=\frac{1}{2}\) and \(c=-1\)A1 OE. \((\ln y = \sqrt{x+1}-1)\). Their value of \(c\) will depend on where \(c\) is in their equation and whether they are working with \(\frac{1}{k}\ln y\). The value of \(k\) must be consistent with what they integrated.
Obtain \(y=\exp(\sqrt{x+1}-1)\)A1 NFWW, OE, ISW.
Total7
## Question 7:

| Answer | Mark | Guidance |
|--------|------|----------|
| State equation $\frac{dy}{dx} = k\frac{y}{\sqrt{x+1}}$ | B1 | OE. Must be a differential equation. |
| Separate variables correctly for *their* differential equation and integrate at least one side | \*M1 | $\int\frac{1}{y}dy = \int\frac{k}{\sqrt{x+1}}dx$ |
| Obtain $\ln y$ | A1 | Allow M1A1A1 if they have assumed $k=1$ or are working with an incorrect value for $k$ |
| Obtain $2[k]\sqrt{x+1}$ | A1 | |
| Use $(0,1)$ **and** $(3,e)$ in an expression containing $\ln y$ and $\sqrt{x+1}$ and a constant of integration to determine $k$ and/or a constant of integration $c$ (or use $(0,1)$, $(3,e)$ and $(x,y)$ as limits on definite integrals) | DM1 | If remove logs before finding the constant of integration then the constant must be of the correct form. |
| Obtain $k=\frac{1}{2}$ and $c=-1$ | A1 | OE. $(\ln y = \sqrt{x+1}-1)$. Their value of $c$ will depend on where $c$ is in their equation and whether they are working with $\frac{1}{k}\ln y$. The value of $k$ must be consistent with what they integrated. |
| Obtain $y=\exp(\sqrt{x+1}-1)$ | A1 | NFWW, OE, ISW. |
| **Total** | **7** | |

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7 A curve is such that the gradient at a general point with coordinates $( x , y )$ is proportional to $\frac { y } { \sqrt { x + 1 } }$. The curve passes through the points with coordinates $( 0,1 )$ and $( 3 , e )$.

By setting up and solving a differential equation, find the equation of the curve, expressing $y$ in terms of $x$.\\

\hfill \mbox{\textit{CAIE P3 2021 Q7 [7]}}