Standard +0.3 This is a straightforward separable differential equation requiring standard techniques: setting up dy/dx = ky²/x, separating variables, integrating both sides (including ln x and -1/y terms), then using two boundary conditions to find constants k and c. While it requires multiple steps and careful algebra, it follows a well-practiced procedure with no conceptual surprises, making it slightly easier than average.
A certain curve is such that its gradient at a general point with coordinates \((x, y)\) is proportional to \(\frac{y^2}{x}\). The curve passes through the points with coordinates \((1, 1)\) and \((e, 2)\). By setting up and solving a differential equation, find the equation of the curve, expressing \(y\) in terms of \(x\). [8]
SC: If k = 1 seen or implied give B0 and then allow B1B1B0M1,
max 3/8.
Answer
Marks
Guidance
Separate variables correctly and integrate at least one side
B1
k 1
∫ dx=∫ dy
x y2
Allow with incorrect value substituted for k
1
Obtain terms – and k ln x
Answer
Marks
Guidance
y
B1 + B1
Incorrect k used scores max. B1B0
Use given coordinates correctly to find k and/or a constant of
integration C in an equation containing terms a , blnx and C
Answer
Marks
Guidance
y
M1
SC: If an incorrect method is used to find k, M1 is allowable for
a correct method to find C
1
Obtain k = and c = – 1, or equivalent
Answer
Marks
Guidance
2
A1 + A1
1 1
lnx=1− A0 for fortuitous answers.
2 y
2
Obtain answer y = , or equivalent, and ISW
Answer
Marks
Guidance
2−lnx
A1
−1
y=
−1+ln x
SC: MR of the fraction.
dy y2
=k B1
dx x2
Separate variables and integrate B1
−1 = −k(+C ) B1+B1
y x
Substitute to find k and/or c M1
e 2−e
k = , c= A1+A1
( ) ( )
2 e−1 2 e−1
Answer A0
8
Answer
Marks
Guidance
Question
Answer
Marks
Question 6:
6 | dy y2
State equation = k , or equivalent
dx x | B1 | SC: If k = 1 seen or implied give B0 and then allow B1B1B0M1,
max 3/8.
Separate variables correctly and integrate at least one side | B1 | k 1
∫ dx=∫ dy
x y2
Allow with incorrect value substituted for k
1
Obtain terms – and k ln x
y | B1 + B1 | Incorrect k used scores max. B1B0
Use given coordinates correctly to find k and/or a constant of
integration C in an equation containing terms a , blnx and C
y | M1 | SC: If an incorrect method is used to find k, M1 is allowable for
a correct method to find C
1
Obtain k = and c = – 1, or equivalent
2 | A1 + A1 | 1 1
lnx=1− A0 for fortuitous answers.
2 y
2
Obtain answer y = , or equivalent, and ISW
2−lnx | A1 | −1
y=
−1+ln x
SC: MR of the fraction.
dy y2
=k B1
dx x2
Separate variables and integrate B1
−1 = −k(+C ) B1+B1
y x
Substitute to find k and/or c M1
e 2−e
k = , c= A1+A1
( ) ( )
2 e−1 2 e−1
Answer A0
8
Question | Answer | Marks | Guidance
A certain curve is such that its gradient at a general point with coordinates $(x, y)$ is proportional to $\frac{y^2}{x}$. The curve passes through the points with coordinates $(1, 1)$ and $(e, 2)$. By setting up and solving a differential equation, find the equation of the curve, expressing $y$ in terms of $x$. [8]
\hfill \mbox{\textit{CAIE P3 2018 Q6 [8]}}